OP Malhotra Class 12 Maths Solutions Chapter 8 Differentiation Exercise 8 (D)

Differentiate w.r.t. x :

Question 1.
(i) log cos x
(ii) log sin x
(iii) cos (log x)
(iv) \( \frac{1}{\log \cos x} \)
(v) x log x - x
Answer:
(i) Let \( y = \log \cos x \)
To find the derivative, we differentiate both sides with respect to x. We use the chain rule here.
\( \frac{dy}{dx} = \frac{1}{\cos x} \frac{d}{dx} (\cos x) \)
\( = \frac{1}{\cos x} (-\sin x) \)
\( = -\frac{\sin x}{\cos x} \)
\( = -\tan x \)
(ii) Let \( y = \log \sin x \)
Differentiating both sides with respect to x, we apply the chain rule.
\( \frac{dy}{dx} = \frac{1}{\sin x} \frac{d}{dx} (\sin x) \)
\( = \frac{1}{\sin x} (\cos x) \)
\( = \frac{\cos x}{\sin x} \)
\( = \cot x \)
(iii) Let \( y = \cos (\log x) \)
Differentiating both sides with respect to x, using the chain rule again.
\( \frac{dy}{dx} = \frac{d}{dx} (\cos (\log x)) \)
\( = -\sin (\log x) \frac{d}{dx} (\log x) \)
\( = -\sin (\log x) \cdot \frac{1}{x} \)
\( = -\frac{\sin (\log x)}{x} \)
(iv) Let \( y = \frac{1}{\log \cos x} \)
We can rewrite this as \( y = (\log \cos x)^{-1} \). Now, we differentiate both sides with respect to x, using the chain rule and power rule.
\( \frac{dy}{dx} = \frac{d}{dx} ((\log \cos x)^{-1}) \)
\( = -1 \cdot (\log \cos x)^{-2} \frac{d}{dx} (\log \cos x) \)
\( = \frac{-1}{(\log \cos x)^2} \cdot \frac{1}{\cos x} \frac{d}{dx} (\cos x) \)
\( = \frac{-1}{(\log \cos x)^2} \cdot \frac{1}{\cos x} (-\sin x) \)
\( = \frac{\sin x}{(\log \cos x)^2 \cos x} \)
\( = \frac{\tan x}{(\log \cos x)^2} \)
(v) Let \( y = x \log x - x \)
Differentiating both sides with respect to x, we use the product rule for \( x \log x \) and differentiate \( -x \).
\( \frac{dy}{dx} = \frac{d}{dx} (x \log x) - \frac{d}{dx} (x) \)
\( = \left( x \cdot \frac{d}{dx} (\log x) + (\log x) \frac{d}{dx} (x) \right) - 1 \)
\( = \left( x \cdot \frac{1}{x} + (\log x) \cdot 1 \right) - 1 \)
\( = (1 + \log x) - 1 \)
\( = \log x \)
In simple words: To differentiate, we find how fast each function changes. For functions inside other functions, like log of cos x, we use the chain rule. This means we differentiate the outer function first, then multiply by the derivative of the inner function. For products, we use the product rule.

🎯 Exam Tip: Always remember the chain rule for composite functions. Break down complex functions into simpler parts and differentiate step-by-step. Pay close attention to negative signs and reciprocal functions.

Question 2. Differentiate the following with respect to x:
(i) log \( \left(\sqrt[5]{x^3}\right) \)
(ii) log (3 - 7x)
(iii) log x³
(iv) log \( \sqrt{x} \)
(v) \( \frac{\sin x}{\log x} \)
Answer:
(i) Let \( y = \log \left(\sqrt[5]{x^3}\right) \)
First, we simplify the expression using logarithm properties: \( \sqrt[5]{x^3} = x^{3/5} \).
So, \( y = \log(x^{3/5}) \)
Using the property \( \log a^b = b \log a \), we get:
\( y = \frac{3}{5} \log x \)
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{3}{5} \frac{d}{dx} (\log x) \)
\( = \frac{3}{5} \cdot \frac{1}{x} \)
\( = \frac{3}{5x} \)
(ii) Let \( y = \log (3 - 7x) \)
Differentiating both sides with respect to x, we use the chain rule:
\( \frac{dy}{dx} = \frac{d}{dx} (\log (3 - 7x)) \)
\( = \frac{1}{(3 - 7x)} \frac{d}{dx} (3 - 7x) \)
\( = \frac{1}{(3 - 7x)} (-7) \)
\( = \frac{-7}{3 - 7x} \)
(iii) Let \( y = \log x^3 \)
First, we simplify using logarithm properties: \( \log x^3 = 3 \log x \).
So, \( y = 3 \log x \)
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = 3 \frac{d}{dx} (\log x) \)
\( = 3 \cdot \frac{1}{x} \)
\( = \frac{3}{x} \)
(iv) Let \( y = \log \sqrt{x} \)
First, simplify the expression: \( \sqrt{x} = x^{1/2} \).
So, \( y = \log (x^{1/2}) \)
Using the property \( \log a^b = b \log a \), we get:
\( y = \frac{1}{2} \log x \)
Now, we differentiate both sides with respect to x:
\( \frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} (\log x) \)
\( = \frac{1}{2} \cdot \frac{1}{x} \)
\( = \frac{1}{2x} \)
(v) Let \( y = \frac{\sin x}{\log x} \)
Differentiating both sides with respect to x, we use the quotient rule: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).
Here, \( u = \sin x \) and \( v = \log x \).
\( \frac{dy}{dx} = \frac{(\log x) \frac{d}{dx} (\sin x) - (\sin x) \frac{d}{dx} (\log x)}{(\log x)^2} \)
\( = \frac{(\log x) (\cos x) - (\sin x) \left( \frac{1}{x} \right)}{(\log x)^2} \)
\( = \frac{x \cos x \log x - \sin x}{x (\log x)^2} \)
In simple words: When we differentiate logarithmic functions, we often use rules to simplify them first, like bringing the power down. Then, we apply the basic differentiation rules. For fractions, we follow the quotient rule, which helps us differentiate functions that are divided by each other.

🎯 Exam Tip: Simplifying logarithmic expressions using properties like \( \log a^b = b \log a \) or \( \log (\frac{a}{b}) = \log a - \log b \) before differentiating can greatly reduce complexity and chances of error. Remember the quotient rule for rational functions.

Question 3. Differentiate log (cosec x - cot x) with respect to x.
Answer:
Let \( y = \log (\operatorname{cosec} x - \cot x) \)
To find \( \frac{dy}{dx} \), we differentiate both sides with respect to x, using the chain rule.
\( \frac{dy}{dx} = \frac{1}{(\operatorname{cosec} x - \cot x)} \frac{d}{dx} (\operatorname{cosec} x - \cot x) \)
First, we find the derivative of \( (\operatorname{cosec} x - \cot x) \):
\( \frac{d}{dx} (\operatorname{cosec} x) = -\operatorname{cosec} x \cot x \)
\( \frac{d}{dx} (\cot x) = -\operatorname{cosec}^2 x \)
So, \( \frac{d}{dx} (\operatorname{cosec} x - \cot x) = -\operatorname{cosec} x \cot x - (-\operatorname{cosec}^2 x) \)
\( = -\operatorname{cosec} x \cot x + \operatorname{cosec}^2 x \)
\( = \operatorname{cosec} x (-\cot x + \operatorname{cosec} x) \)
\( = \operatorname{cosec} x (\operatorname{cosec} x - \cot x) \)
Now, substitute this back into the expression for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{1}{(\operatorname{cosec} x - \cot x)} \cdot \operatorname{cosec} x (\operatorname{cosec} x - \cot x) \)
\( = \operatorname{cosec} x \)
In simple words: We want to find the derivative of log(cosec x - cot x). We use the chain rule: first differentiate the log function, then differentiate the inner part (cosec x - cot x). When we simplify the terms, many parts cancel out, leaving us with a very simple answer.

🎯 Exam Tip: Be mindful of the derivatives of trigonometric functions and their signs. Recognizing common algebraic simplifications after applying differentiation rules can lead to a much simpler final answer, which examiners appreciate.

Question 4. Differentiate sin(log cos x) with respect to x.
Answer:
Let \( y = \sin (\log \cos x) \)
To find \( \frac{dy}{dx} \), we use the chain rule multiple times, working from the outermost function inwards.
\( \frac{dy}{dx} = \frac{d}{dx} (\sin (\log \cos x)) \)
First, differentiate \( \sin(\text{something}) \), which gives \( \cos(\text{something}) \):
\( = \cos (\log \cos x) \frac{d}{dx} (\log \cos x) \)
Next, differentiate \( \log(\text{something}) \), which gives \( \frac{1}{\text{something}} \):
\( = \cos (\log \cos x) \cdot \frac{1}{\cos x} \frac{d}{dx} (\cos x) \)
Finally, differentiate \( \cos x \), which gives \( -\sin x \):
\( = \cos (\log \cos x) \cdot \frac{1}{\cos x} (-\sin x) \)
\( = -\cos (\log \cos x) \cdot \frac{\sin x}{\cos x} \)
\( = -\cos (\log \cos x) \tan x \)
In simple words: This problem involves a function inside a function inside another function. To solve it, we differentiate layer by layer. First, we differentiate the sine part, then the log part, and finally the cosine part, multiplying all the results together.

🎯 Exam Tip: For nested functions, apply the chain rule systematically from the outermost function to the innermost function. This ensures all components are differentiated correctly and multiplied in the right order.

Question 5. Differentiate log (log x) with respect to x.
Answer:
Let \( y = \log (\log x) \)
To find \( \frac{dy}{dx} \), we differentiate both sides with respect to x. We will use the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx} (\log (\log x)) \)
First, we differentiate the outer logarithm function, treating \( \log x \) as a single variable. The derivative of \( \log u \) is \( \frac{1}{u} \).
\( = \frac{1}{\log x} \frac{d}{dx} (\log x) \)
Next, we differentiate the inner function, \( \log x \). The derivative of \( \log x \) is \( \frac{1}{x} \).
\( = \frac{1}{\log x} \cdot \frac{1}{x} \)
\( = \frac{1}{x \log x} \)
In simple words: This is a chain rule problem where we have a log inside another log. We first differentiate the outer log, then multiply that by the derivative of the inner log function. This gives us the final simplified result.

🎯 Exam Tip: The chain rule is essential for nested functions. Remember that \( \frac{d}{dx} (\log f(x)) = \frac{1}{f(x)} f'(x) \). Apply it carefully to avoid common mistakes.

Question 6. Differentiate \( \frac{\log x}{1+\log x} \) with respect to x.
Answer:
Let \( y = \frac{\log x}{1+\log x} \)
To find \( \frac{dy}{dx} \), we differentiate both sides with respect to x. We use the quotient rule, which is for differentiating functions that are divided by each other: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).
Here, \( u = \log x \) and \( v = 1+\log x \).
First, find the derivatives of \( u \) and \( v \):
\( \frac{du}{dx} = \frac{d}{dx} (\log x) = \frac{1}{x} \)
\( \frac{dv}{dx} = \frac{d}{dx} (1+\log x) = 0 + \frac{1}{x} = \frac{1}{x} \)
Now, substitute these into the quotient rule formula:
\[ \frac{dy}{dx} = \frac{(1+\log x) \frac{1}{x} - (\log x) \frac{1}{x}}{(1+\log x)^2} \] Take \( \frac{1}{x} \) common from the numerator:
\[ = \frac{\frac{1}{x} ( (1+\log x) - \log x )}{(1+\log x)^2} \] Simplify the numerator:
\[ = \frac{\frac{1}{x} (1)}{(1+\log x)^2} \] \[ = \frac{1}{x (1+\log x)^2} \]
In simple words: We have a fraction with log x at the top and 1 + log x at the bottom. To differentiate this, we use the quotient rule. We differentiate the top, then the bottom, and combine them according to the rule. After that, we simplify the expression.

🎯 Exam Tip: The quotient rule can be error-prone, especially with complex numerators and denominators. Write down \( u, v, \frac{du}{dx}, \frac{dv}{dx} \) clearly before substituting them into the formula to prevent mistakes.

Question 7. Differentiate sin (log x) - log sin x with respect to x.
Answer:
Let \( y = \sin (\log x) - \log \sin x \)
To find \( \frac{dy}{dx} \), we differentiate each term separately with respect to x. We will use the chain rule for both terms.
\[ \frac{dy}{dx} = \frac{d}{dx} (\sin (\log x)) - \frac{d}{dx} (\log \sin x) \] For the first term, \( \frac{d}{dx} (\sin (\log x)) \):
\( = \cos (\log x) \frac{d}{dx} (\log x) \)
\( = \cos (\log x) \cdot \frac{1}{x} \)
For the second term, \( \frac{d}{dx} (\log \sin x) \):
\( = \frac{1}{\sin x} \frac{d}{dx} (\sin x) \)
\( = \frac{1}{\sin x} (\cos x) \)
\( = \cot x \)
Now, combine the derivatives of both terms:
\[ \frac{dy}{dx} = \frac{\cos (\log x)}{x} - \cot x \]
In simple words: We have two parts to differentiate, separated by a minus sign. For each part, we use the chain rule because a function is inside another function. We differentiate each part carefully and then subtract the results.

🎯 Exam Tip: When differentiating a sum or difference of functions, differentiate each term independently. Be careful with applying the chain rule to each component correctly, especially for trigonometric and logarithmic functions.

Question 8. Differentiate \( \log \left(\frac{1-x^2}{1+x^2}\right) \) with respect to x.
Answer:
Let \( y = \log \left(\frac{1-x^2}{1+x^2}\right) \)
**Method 1: Using the Chain Rule with Quotient Rule**
We differentiate \( \log u \) as \( \frac{1}{u} \frac{du}{dx} \). Here \( u = \frac{1-x^2}{1+x^2} \).
\[ \frac{dy}{dx} = \frac{1}{\left(\frac{1-x^2}{1+x^2}\right)} \frac{d}{dx} \left(\frac{1-x^2}{1+x^2}\right) \] \[ = \frac{1+x^2}{1-x^2} \cdot \frac{(1+x^2)\frac{d}{dx}(1-x^2) - (1-x^2)\frac{d}{dx}(1+x^2)}{(1+x^2)^2} \] We know \( \frac{d}{dx}(1-x^2) = -2x \) and \( \frac{d}{dx}(1+x^2) = 2x \).
\[ = \frac{1+x^2}{1-x^2} \cdot \frac{(1+x^2)(-2x) - (1-x^2)(2x)}{(1+x^2)^2} \] \[ = \frac{1}{1-x^2} \cdot \frac{-2x(1+x^2) - 2x(1-x^2)}{1+x^2} \] \[ = \frac{1}{1-x^2} \cdot \frac{-2x - 2x^3 - 2x + 2x^3}{1+x^2} \] \[ = \frac{1}{1-x^2} \cdot \frac{-4x}{1+x^2} \] \[ = \frac{-4x}{(1-x^2)(1+x^2)} = \frac{-4x}{1-x^4} \] **Method 2: Using Logarithm Properties (simpler)**
We can simplify the expression first using the logarithm property \( \log \left(\frac{a}{b}\right) = \log a - \log b \).
\( y = \log (1-x^2) - \log (1+x^2) \)
Now, we differentiate both sides with respect to x:
\[ \frac{dy}{dx} = \frac{d}{dx} (\log (1-x^2)) - \frac{d}{dx} (\log (1+x^2)) \] For the first term: \( \frac{d}{dx} (\log (1-x^2)) = \frac{1}{1-x^2} \frac{d}{dx} (1-x^2) = \frac{1}{1-x^2} (-2x) = \frac{-2x}{1-x^2} \) For the second term: \( \frac{d}{dx} (\log (1+x^2)) = \frac{1}{1+x^2} \frac{d}{dx} (1+x^2) = \frac{1}{1+x^2} (2x) = \frac{2x}{1+x^2} \) So, combining them:
\[ \frac{dy}{dx} = \frac{-2x}{1-x^2} - \frac{2x}{1+x^2} \] \[ = -2x \left( \frac{1}{1-x^2} + \frac{1}{1+x^2} \right) \] \[ = -2x \left( \frac{(1+x^2) + (1-x^2)}{(1-x^2)(1+x^2)} \right) \] \[ = -2x \left( \frac{1+x^2+1-x^2}{1-x^4} \right) \] \[ = -2x \left( \frac{2}{1-x^4} \right) \] \[ = \frac{-4x}{1-x^4} \]
In simple words: This problem asks us to differentiate a logarithm of a fraction. It's much easier to first use a log rule to turn the fraction into a subtraction of two logs. Then, we differentiate each log term separately using the chain rule, which makes the whole process simpler. Both methods give the same answer.

🎯 Exam Tip: Always look for opportunities to simplify expressions using logarithm properties (like \( \log(A/B) = \log A - \log B \)) *before* differentiating. This often transforms a complicated quotient rule problem into simpler chain rule applications, reducing potential errors.

Question 9. Differentiate \( \log \sqrt{\frac{x-1}{x+1}} \) with respect to x.
Answer:
Let \( y = \log \sqrt{\frac{x-1}{x+1}} \)
First, we simplify the expression using logarithm properties.
We know that \( \sqrt{A} = A^{1/2} \), so \( y = \log \left( \left(\frac{x-1}{x+1}\right)^{1/2} \right) \).
Using the property \( \log a^b = b \log a \):
\( y = \frac{1}{2} \log \left(\frac{x-1}{x+1}\right) \)
Now, using the property \( \log \left(\frac{a}{b}\right) = \log a - \log b \):
\( y = \frac{1}{2} [\log (x-1) - \log (x+1)] \)
Now, we differentiate both sides with respect to x:
\[ \frac{dy}{dx} = \frac{1}{2} \left[ \frac{d}{dx} (\log (x-1)) - \frac{d}{dx} (\log (x+1)) \right] \] For \( \frac{d}{dx} (\log (x-1)) \): Using the chain rule, it's \( \frac{1}{x-1} \frac{d}{dx} (x-1) = \frac{1}{x-1} \cdot 1 = \frac{1}{x-1} \).
For \( \frac{d}{dx} (\log (x+1)) \): Using the chain rule, it's \( \frac{1}{x+1} \frac{d}{dx} (x+1) = \frac{1}{x+1} \cdot 1 = \frac{1}{x+1} \).
Substitute these back:
\[ \frac{dy}{dx} = \frac{1}{2} \left[ \frac{1}{x-1} - \frac{1}{x+1} \right] \] Combine the fractions inside the bracket:
\[ = \frac{1}{2} \left[ \frac{(x+1) - (x-1)}{(x-1)(x+1)} \right] \] \[ = \frac{1}{2} \left[ \frac{x+1-x+1}{x^2-1} \right] \] \[ = \frac{1}{2} \left[ \frac{2}{x^2-1} \right] \] \[ = \frac{1}{x^2-1} \]
In simple words: This problem involves a log of a square root of a fraction. The easiest way to solve it is to first use log rules to get rid of the square root and then the fraction, turning it into a simple subtraction of two log terms. After that, we just differentiate each log term.

🎯 Exam Tip: When dealing with logarithms of roots or fractions, always simplify the expression using logarithm properties before differentiating. This transforms a complex problem into simpler, manageable terms, reducing the likelihood of errors.

Question 10. Differentiate sin [sin (log 3x)] with respect to x.
Answer:
Let \( y = \sin [\sin (\log 3x)] \)
To find \( \frac{dy}{dx} \), we apply the chain rule multiple times, moving from the outermost function to the innermost function.
\[ \frac{dy}{dx} = \frac{d}{dx} (\sin [\sin (\log 3x)]) \] First, differentiate the outermost \( \sin(\text{something}) \), which gives \( \cos(\text{something}) \):
\( = \cos (\sin (\log 3x)) \frac{d}{dx} (\sin (\log 3x)) \)
Next, differentiate the middle \( \sin(\text{something}) \), which gives \( \cos(\text{something}) \):
\( = \cos (\sin (\log 3x)) \cos (\log 3x) \frac{d}{dx} (\log 3x) \)
Finally, differentiate \( \log(\text{something}) \), which gives \( \frac{1}{\text{something}} \), and then differentiate \( 3x \):
\( = \cos (\sin (\log 3x)) \cos (\log 3x) \cdot \frac{1}{3x} \frac{d}{dx} (3x) \)
\( = \cos (\sin (\log 3x)) \cos (\log 3x) \cdot \frac{1}{3x} \cdot 3 \)
The 3 in the numerator and denominator cancel out:
\( = \frac{\cos [\sin (\log 3x)] \cos (\log 3x)}{x} \)
In simple words: We have a very nested function here: sine of (sine of (log of 3x)). We need to peel off each layer of the function one by one, using the chain rule. We differentiate the outermost sine, then the next sine, then the log, and finally the 3x, multiplying all the results together.

🎯 Exam Tip: When dealing with multiple nested functions, systematically apply the chain rule layer by layer. Identify the outermost function, differentiate it, and then multiply by the derivative of its argument, repeating this process until all functions are differentiated.

Question 11. Differentiate log cos \( \sqrt{x} \) with respect to x.
Answer:
Let \( y = \log \cos \sqrt{x} \)
To find \( \frac{dy}{dx} \), we apply the chain rule multiple times.
\[ \frac{dy}{dx} = \frac{d}{dx} (\log \cos \sqrt{x}) \] First, differentiate \( \log(\text{something}) \), which gives \( \frac{1}{\text{something}} \):
\( = \frac{1}{\cos \sqrt{x}} \frac{d}{dx} (\cos \sqrt{x}) \)
Next, differentiate \( \cos(\text{something}) \), which gives \( -\sin(\text{something}) \):
\( = \frac{1}{\cos \sqrt{x}} (-\sin \sqrt{x}) \frac{d}{dx} (\sqrt{x}) \)
Finally, differentiate \( \sqrt{x} \), which is \( x^{1/2} \). The derivative is \( \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \):
\( = \frac{-\sin \sqrt{x}}{\cos \sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \)
We know \( \frac{\sin \sqrt{x}}{\cos \sqrt{x}} = \tan \sqrt{x} \):
\( = -\tan \sqrt{x} \cdot \frac{1}{2\sqrt{x}} \)
\( = -\frac{\tan \sqrt{x}}{2\sqrt{x}} \)
In simple words: This problem asks us to find the derivative of log of cosine of square root x. We use the chain rule three times. First, we differentiate the log function, then the cosine function, and finally the square root function, multiplying all results together.

🎯 Exam Tip: When applying the chain rule, identify each layer of the function and differentiate from the outermost to the innermost. Remember the standard derivatives of log, cos, and square root functions, as incorrect application of these can lead to wrong answers.

Question 12. Differentiate cos \( (\log x)^2 \) with respect to x.
Answer:
Let \( y = \cos (\log x)^2 \)
To find \( \frac{dy}{dx} \), we apply the chain rule layer by layer.
\[ \frac{dy}{dx} = \frac{d}{dx} (\cos (\log x)^2) \] First, differentiate \( \cos(\text{something}) \), which gives \( -\sin(\text{something}) \):
\( = -\sin ((\log x)^2) \frac{d}{dx} ((\log x)^2) \)
Next, differentiate \( (\text{something})^2 \), which gives \( 2 \cdot (\text{something}) \):
\( = -\sin ((\log x)^2) \cdot 2 (\log x) \frac{d}{dx} (\log x) \)
Finally, differentiate \( \log x \), which gives \( \frac{1}{x} \):
\( = -\sin ((\log x)^2) \cdot 2 (\log x) \cdot \frac{1}{x} \)
\[ = \frac{-2 \sin ((\log x)^2) \log x}{x} \]
In simple words: We are differentiating cosine of (log x squared). We apply the chain rule by differentiating the cosine function first. Then, we differentiate the squared term, and finally, we differentiate the log x term, multiplying all these parts together for the final answer.

🎯 Exam Tip: Be careful with the order of operations when applying the chain rule. In \( \cos (\log x)^2 \), the square applies to \( \log x \) as a whole, not just \( x \). Differentiate the outermost function first, then its argument, and so on.

Question 13. Differentiate log \( (\sqrt{\tan x}) \) with respect to x.
Answer:
Let \( y = \log (\sqrt{\tan x}) \)
First, we simplify the expression using logarithm properties. We know \( \sqrt{\tan x} = (\tan x)^{1/2} \).
So, \( y = \log ((\tan x)^{1/2}) \)
Using the property \( \log a^b = b \log a \):
\( y = \frac{1}{2} \log (\tan x) \)
Now, we differentiate both sides with respect to x:
\[ \frac{dy}{dx} = \frac{1}{2} \frac{d}{dx} (\log (\tan x)) \] We use the chain rule for \( \log (\tan x) \). The derivative of \( \log u \) is \( \frac{1}{u} \frac{du}{dx} \).
\( = \frac{1}{2} \cdot \frac{1}{\tan x} \frac{d}{dx} (\tan x) \)
The derivative of \( \tan x \) is \( \sec^2 x \).
\( = \frac{1}{2} \cdot \frac{1}{\tan x} \cdot \sec^2 x \)
We can rewrite \( \tan x = \frac{\sin x}{\cos x} \) and \( \sec^2 x = \frac{1}{\cos^2 x} \):
\( = \frac{1}{2} \cdot \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} \)
\( = \frac{1}{2 \sin x \cos x} \)
We know that \( 2 \sin x \cos x = \sin (2x) \).
\( = \frac{1}{\sin (2x)} \)
\( = \operatorname{cosec} (2x) \)
In simple words: This problem involves differentiating a log of a square root of tan x. It's much easier to first use log rules to simplify the expression by removing the square root. Then, we differentiate using the chain rule and simplify the trigonometric terms.

🎯 Exam Tip: Always simplify logarithmic expressions using properties (like \( \log A^b = b \log A \)) before differentiating. This often reduces the complexity. Also, be prepared to use trigonometric identities (like \( \sin 2x = 2 \sin x \cos x \)) to simplify the final answer.

Question 14. Differentiate \( \log \left(x+\sqrt{1+x^2}\right) \) with respect to x.
Answer:
Let \( y = \log \left(x+\sqrt{1+x^2}\right) \)
To find \( \frac{dy}{dx} \), we use the chain rule.
\[ \frac{dy}{dx} = \frac{d}{dx} \left(\log \left(x+\sqrt{1+x^2}\right)\right) \] The derivative of \( \log u \) is \( \frac{1}{u} \frac{du}{dx} \). Here \( u = x+\sqrt{1+x^2} \).
\[ = \frac{1}{x+\sqrt{1+x^2}} \frac{d}{dx} \left(x+\sqrt{1+x^2}\right) \] Now we need to differentiate \( x+\sqrt{1+x^2} \).
\( \frac{d}{dx} (x) = 1 \)
For \( \frac{d}{dx} (\sqrt{1+x^2}) \), we use the chain rule again.
\( \frac{d}{dx} ((1+x^2)^{1/2}) = \frac{1}{2} (1+x^2)^{-1/2} \frac{d}{dx} (1+x^2) \)
\( = \frac{1}{2\sqrt{1+x^2}} (2x) \)
\( = \frac{x}{\sqrt{1+x^2}} \)
So, \( \frac{d}{dx} \left(x+\sqrt{1+x^2}\right) = 1 + \frac{x}{\sqrt{1+x^2}} \)
Now, substitute this back into the \( \frac{dy}{dx} \) expression:
\[ \frac{dy}{dx} = \frac{1}{x+\sqrt{1+x^2}} \left(1 + \frac{x}{\sqrt{1+x^2}}\right) \] Combine the terms inside the second bracket by finding a common denominator:
\[ = \frac{1}{x+\sqrt{1+x^2}} \left(\frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2}}\right) \] Notice that the term \( (x+\sqrt{1+x^2}) \) in the denominator cancels with the term \( (\sqrt{1+x^2} + x) \) in the numerator.
\[ = \frac{1}{\sqrt{1+x^2}} \]
In simple words: This problem asks for the derivative of a logarithm where the inner part is a sum involving a square root. We use the chain rule: differentiate the log part, then differentiate the inner sum. The inner sum itself requires differentiating a square root, which also uses the chain rule. Finally, we simplify the whole expression.

🎯 Exam Tip: For expressions like \( \sqrt{f(x)} \), remember that its derivative is \( \frac{f'(x)}{2\sqrt{f(x)}} \). This quick derivative can save time. Look for cancellation opportunities after applying the chain rule, as they often simplify the final result significantly.

Question 15. Differentiate \( \log \left(x-\sqrt{x^2-a^2}\right) \) with respect to x.
Answer:
Let \( y = \log \left(x-\sqrt{x^2-a^2}\right) \)
To find \( \frac{dy}{dx} \), we use the chain rule.
\[ \frac{dy}{dx} = \frac{d}{dx} \left(\log \left(x-\sqrt{x^2-a^2}\right)\right) \] The derivative of \( \log u \) is \( \frac{1}{u} \frac{du}{dx} \). Here \( u = x-\sqrt{x^2-a^2} \).
\[ = \frac{1}{x-\sqrt{x^2-a^2}} \frac{d}{dx} \left(x-\sqrt{x^2-a^2}\right) \] Now we need to differentiate \( x-\sqrt{x^2-a^2} \).
\( \frac{d}{dx} (x) = 1 \)
For \( \frac{d}{dx} (\sqrt{x^2-a^2}) \), we use the chain rule:
\( \frac{d}{dx} ((x^2-a^2)^{1/2}) = \frac{1}{2} (x^2-a^2)^{-1/2} \frac{d}{dx} (x^2-a^2) \)
\( = \frac{1}{2\sqrt{x^2-a^2}} (2x) \)
\( = \frac{x}{\sqrt{x^2-a^2}} \)
So, \( \frac{d}{dx} \left(x-\sqrt{x^2-a^2}\right) = 1 - \frac{x}{\sqrt{x^2-a^2}} \)
Now, substitute this back into the \( \frac{dy}{dx} \) expression:
\[ \frac{dy}{dx} = \frac{1}{x-\sqrt{x^2-a^2}} \left(1 - \frac{x}{\sqrt{x^2-a^2}}\right) \] Combine the terms inside the second bracket by finding a common denominator:
\[ = \frac{1}{x-\sqrt{x^2-a^2}} \left(\frac{\sqrt{x^2-a^2} - x}{\sqrt{x^2-a^2}}\right) \] Factor out -1 from the numerator of the second fraction: \( \sqrt{x^2-a^2} - x = -(x - \sqrt{x^2-a^2}) \).
\[ = \frac{1}{x-\sqrt{x^2-a^2}} \left(\frac{-(x - \sqrt{x^2-a^2})}{\sqrt{x^2-a^2}}\right) \] Now, the term \( (x - \sqrt{x^2-a^2}) \) cancels out:
\[ = \frac{-1}{\sqrt{x^2-a^2}} \]
In simple words: We need to differentiate a log function where the inner part is a difference involving a square root. We use the chain rule, differentiating the log first, then the complex term inside. Differentiating the square root part also uses the chain rule. After combining terms and simplifying, many parts cancel, leading to a simple result.

🎯 Exam Tip: Pay close attention to signs, especially when combining fractions. Factoring out negative signs can often reveal opportunities for cancellation, which is crucial for simplifying expressions in differentiation problems.

Question 16. Differentiate \( \sec x \tan x + \log \tan \left(\frac{\pi}{4}+\frac{x}{2}\right) \) with respect to x.
Answer:
Let \( y = \sec x \tan x + \log \tan \left(\frac{\pi}{4}+\frac{x}{2}\right) \)
To find \( \frac{dy}{dx} \), we differentiate each term separately.
\[ \frac{dy}{dx} = \frac{d}{dx} (\sec x \tan x) + \frac{d}{dx} \left(\log \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right) \] **Part 1: Differentiating \( \sec x \tan x \)**
We use the product rule: \( \frac{d}{dx} (uv) = u'v + uv' \). Here \( u = \sec x \) and \( v = \tan x \).
\( \frac{d}{dx} (\sec x) = \sec x \tan x \)
\( \frac{d}{dx} (\tan x) = \sec^2 x \)
So, \( \frac{d}{dx} (\sec x \tan x) = (\sec x \tan x)(\tan x) + (\sec x)(\sec^2 x) \)
\( = \sec x \tan^2 x + \sec^3 x \)
\( = \sec x (\tan^2 x + \sec^2 x) \)
We know \( \tan^2 x = \sec^2 x - 1 \), so \( \tan^2 x + \sec^2 x = \sec^2 x - 1 + \sec^2 x = 2\sec^2 x - 1 \).
\( = \sec x (2\sec^2 x - 1) = 2\sec^3 x - \sec x \)
**Part 2: Differentiating \( \log \tan \left(\frac{\pi}{4}+\frac{x}{2}\right) \)**
We use the chain rule three times.
\( \frac{d}{dx} \left(\log \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right) = \frac{1}{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)} \frac{d}{dx} \left(\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right) \)
\( = \frac{1}{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot \sec^2 \left(\frac{\pi}{4}+\frac{x}{2}\right) \frac{d}{dx} \left(\frac{\pi}{4}+\frac{x}{2}\right) \)
\( = \frac{1}{\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot \sec^2 \left(\frac{\pi}{4}+\frac{x}{2}\right) \cdot \frac{1}{2} \)
Rewrite \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \):
\( = \frac{\cos \left(\frac{\pi}{4}+\frac{x}{2}\right)}{\sin \left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot \frac{1}{\cos^2 \left(\frac{\pi}{4}+\frac{x}{2}\right)} \cdot \frac{1}{2} \)
\( = \frac{1}{2 \sin \left(\frac{\pi}{4}+\frac{x}{2}\right) \cos \left(\frac{\pi}{4}+\frac{x}{2}\right)} \)
Using the identity \( 2 \sin A \cos A = \sin (2A) \):
\( = \frac{1}{\sin \left(2 \left(\frac{\pi}{4}+\frac{x}{2}\right)\right)} = \frac{1}{\sin \left(\frac{\pi}{2}+x\right)} \)
We know that \( \sin \left(\frac{\pi}{2}+x\right) = \cos x \).
\( = \frac{1}{\cos x} = \sec x \)
**Combining Part 1 and Part 2:**
\[ \frac{dy}{dx} = (2\sec^3 x - \sec x) + \sec x \] \[ = 2\sec^3 x \]
In simple words: This problem has two main parts connected by a plus sign, so we differentiate each part separately. The first part uses the product rule for sec x times tan x. The second part is a log function, so we use the chain rule multiple times for log of tan of an angle. After differentiating both parts, we combine and simplify using trigonometric identities.

🎯 Exam Tip: Break down complex expressions into manageable parts. For products, use the product rule. For nested functions, apply the chain rule carefully. Remember trigonometric identities like \( \sec^2 x = 1 + \tan^2 x \) and \( \sin(\frac{\pi}{2} + x) = \cos x \) for simplification.

Question 17. Differentiate \( \log \sqrt{\frac{1+\sin x}{1-\sin x}} \) with respect to x.
Answer:
Let \( y = \log \sqrt{\frac{1+\sin x}{1-\sin x}} \)
First, we simplify the expression using logarithm properties.
We know \( \sqrt{A} = A^{1/2} \), so \( y = \log \left( \left(\frac{1+\sin x}{1-\sin x}\right)^{1/2} \right) \).
Using the property \( \log a^b = b \log a \):
\( y = \frac{1}{2} \log \left(\frac{1+\sin x}{1-\sin x}\right) \)
Now, using the property \( \log \left(\frac{a}{b}\right) = \log a - \log b \):
\( y = \frac{1}{2} [\log (1+\sin x) - \log (1-\sin x)] \)
Now, we differentiate both sides with respect to x:
\[ \frac{dy}{dx} = \frac{1}{2} \left[ \frac{d}{dx} (\log (1+\sin x)) - \frac{d}{dx} (\log (1-\sin x)) \right] \] For \( \frac{d}{dx} (\log (1+\sin x)) \): Using the chain rule, it's \( \frac{1}{1+\sin x} \frac{d}{dx} (1+\sin x) = \frac{1}{1+\sin x} (\cos x) = \frac{\cos x}{1+\sin x} \).
For \( \frac{d}{dx} (\log (1-\sin x)) \): Using the chain rule, it's \( \frac{1}{1-\sin x} \frac{d}{dx} (1-\sin x) = \frac{1}{1-\sin x} (-\cos x) = \frac{-\cos x}{1-\sin x} \).
Substitute these back:
\[ \frac{dy}{dx} = \frac{1}{2} \left[ \frac{\cos x}{1+\sin x} - \left(\frac{-\cos x}{1-\sin x}\right) \right] \] \[ = \frac{1}{2} \left[ \frac{\cos x}{1+\sin x} + \frac{\cos x}{1-\sin x} \right] \] Combine the fractions inside the bracket by finding a common denominator:
\[ = \frac{1}{2} \left[ \frac{\cos x (1-\sin x) + \cos x (1+\sin x)}{(1+\sin x)(1-\sin x)} \right] \] \[ = \frac{1}{2} \left[ \frac{\cos x - \cos x \sin x + \cos x + \cos x \sin x}{1-\sin^2 x} \right] \] \[ = \frac{1}{2} \left[ \frac{2 \cos x}{\cos^2 x} \right] \] \[ = \frac{1}{\cos x} \]
\[ = \sec x \]
In simple words: To differentiate this log of a square root of a fraction, we first use log rules to simplify it into a subtraction of two simple log terms. Then we differentiate each log term using the chain rule. After combining the fractions, many terms cancel out, leading to a simple answer.

🎯 Exam Tip: Simplifying complex logarithmic expressions using properties before differentiation is key. Recognizing and applying trigonometric identities like \( \cos^2 x + \sin^2 x = 1 \) (which means \( 1-\sin^2 x = \cos^2 x \)) is vital for simplifying the final result.

Question 18. Differentiate \( \log\sqrt[3]{\frac{1-x}{1+x}} \) with respect to x.
Answer:
Let \( y = \log\sqrt[3]{\frac{1-x}{1+x}} \)
First, we simplify the expression using logarithm properties.
We know \( \sqrt[3]{A} = A^{1/3} \), so \( y = \log \left( \left(\frac{1-x}{1+x}\right)^{1/3} \right) \).
Using the property \( \log a^b = b \log a \):
\( y = \frac{1}{3} \log \left(\frac{1-x}{1+x}\right) \)
Now, using the property \( \log \left(\frac{a}{b}\right) = \log a - \log b \):
\( y = \frac{1}{3} [\log (1-x) - \log (1+x)] \)
Now, we differentiate both sides with respect to x:
\[ \frac{dy}{dx} = \frac{1}{3} \left[ \frac{d}{dx} (\log (1-x)) - \frac{d}{dx} (\log (1+x)) \right] \] For \( \frac{d}{dx} (\log (1-x)) \): Using the chain rule, it's \( \frac{1}{1-x} \frac{d}{dx} (1-x) = \frac{1}{1-x} (-1) = \frac{-1}{1-x} \).
For \( \frac{d}{dx} (\log (1+x)) \): Using the chain rule, it's \( \frac{1}{1+x} \frac{d}{dx} (1+x) = \frac{1}{1+x} (1) = \frac{1}{1+x} \).
Substitute these back:
\[ \frac{dy}{dx} = \frac{1}{3} \left[ \frac{-1}{1-x} - \frac{1}{1+x} \right] \] Combine the fractions inside the bracket by finding a common denominator:
\[ = \frac{1}{3} \left[ \frac{-(1+x) - (1-x)}{(1-x)(1+x)} \right] \] \[ = \frac{1}{3} \left[ \frac{-1-x-1+x}{1-x^2} \right] \] \[ = \frac{1}{3} \left[ \frac{-2}{1-x^2} \right] \] \[ = \frac{-2}{3(1-x^2)} \]
In simple words: To find the derivative of this log function with a cube root and a fraction, we first simplify it using logarithm rules. This turns it into a simple subtraction of two log terms. Then, we differentiate each log term using the chain rule, being careful with signs, and combine the results.

🎯 Exam Tip: Remember to apply the chain rule correctly when differentiating terms like \( \log (1-x) \), where the derivative of the inner function \( (1-x) \) is \( -1 \). Also, a common mistake is to forget the constant factor \( \frac{1}{3} \) throughout the calculation.

Question 19. Differentiate \( (\ln \ln x)^2 \) with respect to x.
Answer:
Let \( y = (\ln (\ln x))^2 \)
To find \( \frac{dy}{dx} \), we apply the chain rule multiple times, moving from the outermost function inwards.
\[ \frac{dy}{dx} = \frac{d}{dx} ((\ln (\ln x))^2) \] First, differentiate \( (\text{something})^2 \), which gives \( 2 \cdot (\text{something}) \):
\( = 2 (\ln (\ln x)) \frac{d}{dx} (\ln (\ln x)) \)
Next, differentiate the outer \( \ln(\text{something}) \), which gives \( \frac{1}{\text{something}} \):
\( = 2 (\ln (\ln x)) \cdot \frac{1}{\ln x} \frac{d}{dx} (\ln x) \)
Finally, differentiate the innermost \( \ln x \), which gives \( \frac{1}{x} \):
\( = 2 (\ln (\ln x)) \cdot \frac{1}{\ln x} \cdot \frac{1}{x} \)
\[ = \frac{2 \ln (\ln x)}{x \ln x} \]
In simple words: This problem involves differentiating a squared term, where the term itself is a natural logarithm of another natural logarithm. We apply the chain rule three times, peeling back the layers: first the power, then the outer natural log, and finally the inner natural log.

🎯 Exam Tip: For nested functions like this, it's crucial to identify the layers of differentiation: power rule first, then outer logarithm, then inner logarithm. Write down each step clearly to avoid missing any part of the chain rule.

Question 20. Differentiate \( \log\left(\sec \frac{x}{2}+\tan \frac{x}{2}\right) \) with respect to x.
Answer:
Let \( y = \log\left(\sec \frac{x}{2}+\tan \frac{x}{2}\right) \)
To find \( \frac{dy}{dx} \), we use the chain rule.
\[ \frac{dy}{dx} = \frac{d}{dx} \left(\log\left(\sec \frac{x}{2}+\tan \frac{x}{2}\right)\right) \] First, differentiate \( \log(\text{something}) \), which gives \( \frac{1}{\text{something}} \):
\[ = \frac{1}{\sec \frac{x}{2}+\tan \frac{x}{2}} \frac{d}{dx} \left(\sec \frac{x}{2}+\tan \frac{x}{2}\right) \] Now, we need to differentiate the term \( \left(\sec \frac{x}{2}+\tan \frac{x}{2}\right) \). We differentiate each part separately.
For \( \frac{d}{dx} \left(\sec \frac{x}{2}\right) \): Using the chain rule, derivative of \( \sec u \) is \( \sec u \tan u \cdot \frac{du}{dx} \).
\( = \sec \frac{x}{2} \tan \frac{x}{2} \cdot \frac{d}{dx} \left(\frac{x}{2}\right) = \sec \frac{x}{2} \tan \frac{x}{2} \cdot \frac{1}{2} \) For \( \frac{d}{dx} \left(\tan \frac{x}{2}\right) \): Using the chain rule, derivative of \( \tan u \) is \( \sec^2 u \cdot \frac{du}{dx} \).
\( = \sec^2 \frac{x}{2} \cdot \frac{d}{dx} \left(\frac{x}{2}\right) = \sec^2 \frac{x}{2} \cdot \frac{1}{2} \) So, \( \frac{d}{dx} \left(\sec \frac{x}{2}+\tan \frac{x}{2}\right) = \frac{1}{2} \sec \frac{x}{2} \tan \frac{x}{2} + \frac{1}{2} \sec^2 \frac{x}{2} \)
Factor out \( \frac{1}{2} \sec \frac{x}{2} \):
\( = \frac{1}{2} \sec \frac{x}{2} \left(\tan \frac{x}{2} + \sec \frac{x}{2}\right) \)
Now, substitute this back into the \( \frac{dy}{dx} \) expression:
\[ \frac{dy}{dx} = \frac{1}{\sec \frac{x}{2}+\tan \frac{x}{2}} \cdot \frac{1}{2} \sec \frac{x}{2} \left(\tan \frac{x}{2} + \sec \frac{x}{2}\right) \] Notice that the term \( \left(\sec \frac{x}{2}+\tan \frac{x}{2}\right) \) in the denominator cancels with the term \( \left(\tan \frac{x}{2} + \sec \frac{x}{2}\right) \) in the numerator.
\[ = \frac{1}{2} \sec \frac{x}{2} \]
In simple words: This problem asks for the derivative of a logarithm of a sum of trigonometric functions. We use the chain rule: first differentiate the log function, then differentiate the sum of secant and tangent. Each of these also requires the chain rule because of \( \frac{x}{2} \). After differentiating, we simplify the terms, and many parts cancel out.

🎯 Exam Tip: For expressions involving \( \frac{x}{2} \) or other linear functions of x inside trigonometric or exponential terms, remember to multiply by the derivative of that inner linear function. Simplification using common factors after differentiation is often a key step to the final answer.

Question 21. Differentiate log [sin (log x)] with respect to x.
Answer:
Let \( y = \log [\sin (\log x)] \)
To find \( \frac{dy}{dx} \), we apply the chain rule multiple times, from the outermost function inwards.
\[ \frac{dy}{dx} = \frac{d}{dx} (\log [\sin (\log x)]) \] First, differentiate the outermost \( \log(\text{something}) \), which gives \( \frac{1}{\text{something}} \):
\( = \frac{1}{\sin (\log x)} \frac{d}{dx} (\sin (\log x)) \)
Next, differentiate \( \sin(\text{something}) \), which gives \( \cos(\text{something}) \):
\( = \frac{1}{\sin (\log x)} \cdot \cos (\log x) \frac{d}{dx} (\log x) \)
Finally, differentiate \( \log x \), which gives \( \frac{1}{x} \):
\( = \frac{\cos (\log x)}{\sin (\log x)} \cdot \frac{1}{x} \)
We know that \( \frac{\cos \theta}{\sin \theta} = \cot \theta \).
\( = \frac{\cot (\log x)}{x} \)
In simple words: This problem involves differentiating a log function that contains a sine function, which then contains another log function. We use the chain rule three times: differentiate the outer log, then the sine, then the inner log. After that, we use a basic trig identity to simplify the answer.

🎯 Exam Tip: Identify the layered structure of the function clearly. Differentiate each layer (log, sin, log) sequentially and multiply the results. Remembering trigonometric identities like \( \frac{\cos x}{\sin x} = \cot x \) is crucial for presenting the answer in its most simplified form.

Question 22. Differentiate \( \log\left[\log \left(\sin \sqrt{x^2+1}\right)\right] \) with respect to x.
Answer:
Let \( y = \log\left[\log \left(\sin \sqrt{x^2+1}\right)\right] \)
To find \( \frac{dy}{dx} \), we apply the chain rule multiple times, moving from the outermost function inwards.
\[ \frac{dy}{dx} = \frac{d}{dx} \left(\log\left[\log \left(\sin \sqrt{x^2+1}\right)\right]\right) \] 1. Differentiate the outermost \( \log(\text{something}) \):
\( = \frac{1}{\log \left(\sin \sqrt{x^2+1}\right)} \frac{d}{dx} \left(\log \left(\sin \sqrt{x^2+1}\right)\right) \)
2. Differentiate the next \( \log(\text{something}) \):
\( = \frac{1}{\log \left(\sin \sqrt{x^2+1}\right)} \cdot \frac{1}{\sin \sqrt{x^2+1}} \frac{d}{dx} \left(\sin \sqrt{x^2+1}\right) \)
3. Differentiate \( \sin(\text{something}) \):
\( = \frac{1}{\log \left(\sin \sqrt{x^2+1}\right) \sin \sqrt{x^2+1}} \cdot \cos \sqrt{x^2+1} \frac{d}{dx} \left(\sqrt{x^2+1}\right) \)
4. Differentiate \( \sqrt{x^2+1} \):
\( \frac{d}{dx} ((x^2+1)^{1/2}) = \frac{1}{2} (x^2+1)^{-1/2} \frac{d}{dx} (x^2+1) \)
\( = \frac{1}{2\sqrt{x^2+1}} (2x) = \frac{x}{\sqrt{x^2+1}} \)
Substitute this back:
\( = \frac{1}{\log \left(\sin \sqrt{x^2+1}\right) \sin \sqrt{x^2+1}} \cdot \cos \sqrt{x^2+1} \cdot \frac{x}{\sqrt{x^2+1}} \)
Rearrange and use \( \frac{\cos \theta}{\sin \theta} = \cot \theta \):
\[ = \frac{x \cot \sqrt{x^2+1}}{\sqrt{x^2+1} \log \left(\sin \sqrt{x^2+1}\right)} \]
In simple words: This function is deeply nested, with a log of a log of a sine of a square root. We use the chain rule repeatedly, differentiating each layer from the outside to the inside. We differentiate the first log, then the second log, then the sine, and finally the square root function, multiplying all the derivatives together.

🎯 Exam Tip: For extremely nested functions, clearly label each step of the chain rule. This prevents errors in complex differentiation. Also, be aware of trigonometric identities like \( \cot \theta = \frac{\cos \theta}{\sin \theta} \) for simplifying the final expression.

Question 23. Differentiate \( \log\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}} \) with respect to x.
Answer:
Let \( y = \log\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}} \)
First, we simplify the expression inside the logarithm by multiplying the numerator and denominator by the conjugate of the denominator, \( (\sqrt{a+x}+\sqrt{a-x}) \).
\[ y = \log\left( \frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}} \times \frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}} \right) \] \[ y = \log\left( \frac{(\sqrt{a+x}+\sqrt{a-x})^2}{(\sqrt{a+x})^2 - (\sqrt{a-x})^2} \right) \] \[ y = \log\left( \frac{(a+x) + (a-x) + 2\sqrt{a+x}\sqrt{a-x}}{(a+x) - (a-x)} \right) \] \[ y = \log\left( \frac{2a + 2\sqrt{(a+x)(a-x)}}{a+x-a+x} \right) \] \[ y = \log\left( \frac{2a + 2\sqrt{a^2-x^2}}{2x} \right) \] \[ y = \log\left( \frac{a + \sqrt{a^2-x^2}}{x} \right) \] Now, using the logarithm property \( \log \left(\frac{A}{B}\right) = \log A - \log B \):
\( y = \log (a + \sqrt{a^2-x^2}) - \log x \)
Now, we differentiate both sides with respect to x:
\[ \frac{dy}{dx} = \frac{d}{dx} (\log (a + \sqrt{a^2-x^2})) - \frac{d}{dx} (\log x) \] For the second term, \( \frac{d}{dx} (\log x) = \frac{1}{x} \).
For the first term, \( \frac{d}{dx} (\log (a + \sqrt{a^2-x^2})) \): We use the chain rule.
\( = \frac{1}{a + \sqrt{a^2-x^2}} \frac{d}{dx} (a + \sqrt{a^2-x^2}) \)
\( = \frac{1}{a + \sqrt{a^2-x^2}} \left( 0 + \frac{d}{dx} (\sqrt{a^2-x^2}) \right) \)
To differentiate \( \sqrt{a^2-x^2} \):
\( \frac{d}{dx} ((a^2-x^2)^{1/2}) = \frac{1}{2} (a^2-x^2)^{-1/2} \frac{d}{dx} (a^2-x^2) \)
\( = \frac{1}{2\sqrt{a^2-x^2}} (-2x) = \frac{-x}{\sqrt{a^2-x^2}} \)
So, the first term's derivative is:
\( = \frac{1}{a + \sqrt{a^2-x^2}} \cdot \frac{-x}{\sqrt{a^2-x^2}} \)
Now, combine both derivatives:
\[ \frac{dy}{dx} = \frac{-x}{(a + \sqrt{a^2-x^2})\sqrt{a^2-x^2}} - \frac{1}{x} \] Find a common denominator to combine:
\[ = \frac{-x^2 - (a + \sqrt{a^2-x^2})\sqrt{a^2-x^2}}{x(a + \sqrt{a^2-x^2})\sqrt{a^2-x^2}} \] Expand the numerator:
\[ = \frac{-x^2 - a\sqrt{a^2-x^2} - (a^2-x^2)}{x(a + \sqrt{a^2-x^2})\sqrt{a^2-x^2}} \] \[ = \frac{-x^2 - a\sqrt{a^2-x^2} - a^2 + x^2}{x(a + \sqrt{a^2-x^2})\sqrt{a^2-x^2}} \] \[ = \frac{-a^2 - a\sqrt{a^2-x^2}}{x(a + \sqrt{a^2-x^2})\sqrt{a^2-x^2}} \] Factor out \( -a \) from the numerator:
\[ = \frac{-a (a + \sqrt{a^2-x^2})}{x(a + \sqrt{a^2-x^2})\sqrt{a^2-x^2}} \] Cancel the common term \( (a + \sqrt{a^2-x^2}) \):
\[ = \frac{-a}{x\sqrt{a^2-x^2}} \]
In simple words: We have a complex logarithm of a fraction. The first step is to simplify the fraction inside the log by multiplying the top and bottom by the conjugate. This simplifies the expression greatly. Then, we use log properties to separate it into two simpler log terms. Finally, we differentiate each term using the chain rule and combine them.

🎯 Exam Tip: When faced with a logarithm of a complex fraction involving square roots, rationalize the expression inside the logarithm first. This transformation simplifies the problem significantly before any differentiation rules need to be applied. Logarithm properties are powerful simplification tools.

Question 24. If \( y = \sqrt{x^2+1}-\log \left(\frac{1}{x}+\sqrt{1+\frac{1}{x^2}}\right) \), find \( \frac{dy}{dx} \).
Answer:
Let \( y = \sqrt{x^2+1}-\log \left(\frac{1}{x}+\sqrt{1+\frac{1}{x^2}}\right) \)
First, let's simplify the term inside the logarithm:
\( \frac{1}{x}+\sqrt{1+\frac{1}{x^2}} = \frac{1}{x}+\sqrt{\frac{x^2+1}{x^2}} = \frac{1}{x}+\frac{\sqrt{x^2+1}}{\sqrt{x^2}} = \frac{1}{x}+\frac{\sqrt{x^2+1}}{x} \) (assuming \( x>0 \))
\( = \frac{1+\sqrt{x^2+1}}{x} \)
So, \( y = \sqrt{x^2+1}-\log \left(\frac{1+\sqrt{x^2+1}}{x}\right) \)
Now, use the logarithm property \( \log \left(\frac{A}{B}\right) = \log A - \log B \):
\( y = \sqrt{x^2+1} - [\log (1+\sqrt{x^2+1}) - \log x] \)
\( y = \sqrt{x^2+1} - \log (1+\sqrt{x^2+1}) + \log x \)
Now, we differentiate each term with respect to x:
\[ \frac{dy}{dx} = \frac{d}{dx} (\sqrt{x^2+1}) - \frac{d}{dx} (\log (1+\sqrt{x^2+1})) + \frac{d}{dx} (\log x) \] 1. **Differentiating \( \sqrt{x^2+1} \):**
\( \frac{d}{dx} ((x^2+1)^{1/2}) = \frac{1}{2} (x^2+1)^{-1/2} \frac{d}{dx} (x^2+1) = \frac{1}{2\sqrt{x^2+1}} (2x) = \frac{x}{\sqrt{x^2+1}} \) 2. **Differentiating \( \log (1+\sqrt{x^2+1}) \):**
\( = \frac{1}{1+\sqrt{x^2+1}} \frac{d}{dx} (1+\sqrt{x^2+1}) \) \( = \frac{1}{1+\sqrt{x^2+1}} \left(0 + \frac{x}{\sqrt{x^2+1}}\right) \) (using the result from step 1 for \( \frac{d}{dx} (\sqrt{x^2+1}) \))
\( = \frac{x}{\sqrt{x^2+1}(1+\sqrt{x^2+1})} \) 3. **Differentiating \( \log x \):**
\( = \frac{1}{x} \)
Now, substitute these three results back into the \( \frac{dy}{dx} \) expression:
\[ \frac{dy}{dx} = \frac{x}{\sqrt{x^2+1}} - \frac{x}{\sqrt{x^2+1}(1+\sqrt{x^2+1})} + \frac{1}{x} \] Combine the first two terms by finding a common denominator:
\[ = \frac{x(1+\sqrt{x^2+1}) - x}{\sqrt{x^2+1}(1+\sqrt{x^2+1})} + \frac{1}{x} \] \[ = \frac{x+x\sqrt{x^2+1} - x}{\sqrt{x^2+1}(1+\sqrt{x^2+1})} + \frac{1}{x} \] \[ = \frac{x\sqrt{x^2+1}}{\sqrt{x^2+1}(1+\sqrt{x^2+1})} + \frac{1}{x} \] The \( \sqrt{x^2+1} \) term cancels out:
\[ = \frac{x}{1+\sqrt{x^2+1}} + \frac{1}{x} \] Now, find a common denominator for these two terms:
\[ = \frac{x^2 + (1+\sqrt{x^2+1})}{x(1+\sqrt{x^2+1})} \] \[ = \frac{x^2 + 1 + \sqrt{x^2+1}}{x(1+\sqrt{x^2+1})} \] This can be rewritten to simplify further. Multiply the first term \( \frac{x}{1+\sqrt{x^2+1}} \) by \( \frac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}-1} \):
\( \frac{x(\sqrt{x^2+1}-1)}{(1+\sqrt{x^2+1})(\sqrt{x^2+1}-1)} = \frac{x(\sqrt{x^2+1}-1)}{(x^2+1)-1} = \frac{x(\sqrt{x^2+1}-1)}{x^2} = \frac{\sqrt{x^2+1}-1}{x} \)
So, the expression becomes:
\[ \frac{dy}{dx} = \frac{\sqrt{x^2+1}-1}{x} + \frac{1}{x} \] \[ = \frac{\sqrt{x^2+1}-1+1}{x} \] \[ = \frac{\sqrt{x^2+1}}{x} \]
In simple words: This problem involves differentiating a complex expression with a square root and a logarithm. First, we simplify the term inside the logarithm. Then, we use logarithm properties to expand the expression into simpler terms. After that, we differentiate each term separately using the chain rule and combine all the results.

🎯 Exam Tip: Simplifying the expression before differentiation, especially inside logarithms, is crucial. For this type of problem, rationalizing the inner part of the logarithm and then using logarithm properties can turn a seemingly difficult differentiation into a series of manageable steps. Be meticulous with algebraic simplification.

Question 23. Differentiate \( \log \left(\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}\right) \) with respect to x.
Answer:
Let the given function be \( y \).
So, \( y = \log \left(\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}\right) \)
First, we simplify the expression inside the logarithm by multiplying the numerator and denominator by \( (\sqrt{a+x}+\sqrt{a-x}) \).
\( y = \log \left(\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}} \times \frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}}\right) \)
\( \implies y = \log \left(\frac{(\sqrt{a+x}+\sqrt{a-x})^2}{(\sqrt{a+x})^2 - (\sqrt{a-x})^2}\right) \)
\( \implies y = \log \left(\frac{(a+x) + (a-x) + 2\sqrt{(a+x)(a-x)}}{(a+x) - (a-x)}\right) \)
\( \implies y = \log \left(\frac{2a + 2\sqrt{a^2-x^2}}{2x}\right) \)
\( \implies y = \log \left(\frac{a + \sqrt{a^2-x^2}}{x}\right) \)
Now, we can use the logarithm property \( \log\left(\frac{A}{B}\right) = \log A - \log B \).
\( y = \log(a + \sqrt{a^2-x^2}) - \log x \)
Next, we differentiate both sides with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx}[\log(a + \sqrt{a^2-x^2}) - \log x] \)
\( \implies \frac{dy}{dx} = \frac{1}{a + \sqrt{a^2-x^2}} \cdot \frac{d}{dx}(a + \sqrt{a^2-x^2}) - \frac{1}{x} \)
\( \implies \frac{dy}{dx} = \frac{1}{a + \sqrt{a^2-x^2}} \cdot \left(0 + \frac{1}{2\sqrt{a^2-x^2}} \cdot (-2x)\right) - \frac{1}{x} \)
\( \implies \frac{dy}{dx} = \frac{1}{a + \sqrt{a^2-x^2}} \cdot \frac{-x}{\sqrt{a^2-x^2}} - \frac{1}{x} \)
\( \implies \frac{dy}{dx} = \frac{-x}{(a + \sqrt{a^2-x^2})\sqrt{a^2-x^2}} - \frac{1}{x} \)
To combine these, find a common denominator.
Common Denominator: \( x(a + \sqrt{a^2-x^2})\sqrt{a^2-x^2} \)
Numerator: \( -x^2 - (a + \sqrt{a^2-x^2})\sqrt{a^2-x^2} \)
\( = -x^2 - (a\sqrt{a^2-x^2} + a^2 - x^2) \)
\( = -x^2 - a\sqrt{a^2-x^2} - a^2 + x^2 \)
\( = -a\sqrt{a^2-x^2} - a^2 \)
\( = -a(a + \sqrt{a^2-x^2}) \)
So, \( \frac{dy}{dx} = \frac{-a(a + \sqrt{a^2-x^2})}{x(a + \sqrt{a^2-x^2})\sqrt{a^2-x^2}} \)
Cancel out the common term \( (a + \sqrt{a^2-x^2}) \).
\( \implies \frac{dy}{dx} = \frac{-a}{x\sqrt{a^2-x^2}} \)
In simple words: First, we made the fraction inside the logarithm simpler by multiplying the top and bottom. This helped us get rid of the square root in the denominator. Then, we used a logarithm rule to split the expression into two easier parts. After that, we took the derivative of each part separately and combined them to get the final answer.

🎯 Exam Tip: Always simplify logarithmic expressions using properties like \( \log(A/B) = \log A - \log B \) or \( \log(A \cdot B) = \log A + \log B \) before differentiating. This often makes the differentiation process much easier and reduces chances of errors.

Question 24. If \( y = \sqrt{x^2+1}-\log \left(\frac{1}{x}+\sqrt{1+\frac{1}{x^2}}\right) \), find \( \frac { dy }{ dx } \).
Answer:
Let the given function be \( y \).
\( y = \sqrt{x^2+1}-\log \left(\frac{1}{x}+\sqrt{1+\frac{1}{x^2}}\right) \)
First, we simplify the expression inside the logarithm.
The term \( \left(\frac{1}{x}+\sqrt{1+\frac{1}{x^2}}\right) \) can be written as:
\( \frac{1}{x}+\sqrt{\frac{x^2+1}{x^2}} \)
\( = \frac{1}{x}+\frac{\sqrt{x^2+1}}{x} \)
\( = \frac{1+\sqrt{x^2+1}}{x} \)
So, the function becomes:
\( y = \sqrt{x^2+1} - \log \left(\frac{1+\sqrt{x^2+1}}{x}\right) \)
Using the logarithm property \( \log\left(\frac{A}{B}\right) = \log A - \log B \):
\( y = \sqrt{x^2+1} - [\log(1+\sqrt{x^2+1}) - \log x] \)
\( \implies y = \sqrt{x^2+1} - \log(1+\sqrt{x^2+1}) + \log x \)
Now, we differentiate each term with respect to x.
\( \frac{dy}{dx} = \frac{d}{dx}(\sqrt{x^2+1}) - \frac{d}{dx}(\log(1+\sqrt{x^2+1})) + \frac{d}{dx}(\log x) \)
Differentiating \( \sqrt{x^2+1} \):
\( \frac{d}{dx}(\sqrt{x^2+1}) = \frac{1}{2\sqrt{x^2+1}} \cdot (2x) = \frac{x}{\sqrt{x^2+1}} \)
Differentiating \( \log(1+\sqrt{x^2+1}) \):
\( \frac{d}{dx}(\log(1+\sqrt{x^2+1})) = \frac{1}{1+\sqrt{x^2+1}} \cdot \frac{d}{dx}(1+\sqrt{x^2+1}) \)
\( = \frac{1}{1+\sqrt{x^2+1}} \cdot \left(0 + \frac{1}{2\sqrt{x^2+1}} \cdot 2x\right) \)
\( = \frac{1}{1+\sqrt{x^2+1}} \cdot \frac{x}{\sqrt{x^2+1}} \)
Differentiating \( \log x \):
\( \frac{d}{dx}(\log x) = \frac{1}{x} \)
Combining these results:
\( \frac{dy}{dx} = \frac{x}{\sqrt{x^2+1}} - \frac{x}{(1+\sqrt{x^2+1})\sqrt{x^2+1}} + \frac{1}{x} \)
Factor out \( \frac{x}{\sqrt{x^2+1}} \) from the first two terms:
\( \frac{dy}{dx} = \frac{x}{\sqrt{x^2+1}} \left(1 - \frac{1}{1+\sqrt{x^2+1}}\right) + \frac{1}{x} \)
\( \implies \frac{dy}{dx} = \frac{x}{\sqrt{x^2+1}} \left(\frac{1+\sqrt{x^2+1}-1}{1+\sqrt{x^2+1}}\right) + \frac{1}{x} \)
\( \implies \frac{dy}{dx} = \frac{x}{\sqrt{x^2+1}} \left(\frac{\sqrt{x^2+1}}{1+\sqrt{x^2+1}}\right) + \frac{1}{x} \)
\( \implies \frac{dy}{dx} = \frac{x}{1+\sqrt{x^2+1}} + \frac{1}{x} \)
In simple words: First, we simplified the complex fraction inside the logarithm by combining terms. Then, we used log rules to break the problem into easier parts. After that, we found the derivative of each part separately and added them up to get the final answer.

🎯 Exam Tip: For complex functions involving logarithms and square roots, always simplify the expression as much as possible using algebraic manipulations and logarithm properties BEFORE attempting to differentiate. This will prevent common errors and make the process more manageable.