GSEB Class 12 Maths Solutions Chapter 5 Continuity and Differentiability Exercise 5.4

Question 1. Find the derivative of \( \frac{e^{x}}{\sin x} \).
Answer: Let the function be \( y = \frac{e^{x}}{\sin x} \). We use the quotient rule for differentiation, which states that if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{u'v-uv'}{v^2} \).
Here, let \( u = e^x \) and \( v = \sin x \).
First, we find the derivatives of \( u \) and \( v \):
The derivative of \( u \) with respect to \( x \) is \( u' = \frac{d}{dx}(e^x) = e^x \).
The derivative of \( v \) with respect to \( x \) is \( v' = \frac{d}{dx}(\sin x) = \cos x \).
Now, we apply the quotient rule:
\( \frac{dy}{dx} = \frac{(e^x)(\sin x) - (e^x)(\cos x)}{(\sin x)^2} \)
We can factor out \( e^x \) from the numerator:

\( \implies \frac{dy}{dx} = \frac{e^x (\sin x - \cos x)}{\sin^2 x} \)
This derivative is valid for all \( x \) where \( \sin x \neq 0 \), which means \( x \neq n\pi \), where \( n \) is an integer.
In simple words: To differentiate a fraction, you use a special rule. You take the derivative of the top part, multiply by the bottom part, then subtract the top part times the derivative of the bottom part. All of this is divided by the bottom part squared.

Exam Tip: Always remember the quotient rule formula accurately, especially the order of terms and the subtraction. Factoring out common terms like \( e^x \) can help simplify the final answer.

Question 2. Find the derivative of \( e^{\sin^{-1}x} \).
Answer: Let the given function be \( y = e^{\sin^{-1}x} \). To differentiate this, we use the chain rule because it's a composite function (a function within a function).
Let \( t = \sin^{-1}x \).
Then the expression becomes \( y = e^t \).
Now, we find the derivatives of \( y \) with respect to \( t \) and \( t \) with respect to \( x \):
The derivative of \( y = e^t \) with respect to \( t \) is \( \frac{dy}{dt} = e^t \).
The derivative of \( t = \sin^{-1}x \) with respect to \( x \) is \( \frac{dt}{dx} = \frac{1}{\sqrt{1-x^2}} \).
According to the chain rule, \( \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} \).
Substituting the derivatives we just found:
\( \frac{dy}{dx} = e^t \cdot \frac{1}{\sqrt{1-x^2}} \)
Finally, substitute \( t = \sin^{-1}x \) back into the expression:

\( \implies \frac{dy}{dx} = e^{\sin^{-1}x} \cdot \frac{1}{\sqrt{1-x^2}} \)

\( \implies \frac{dy}{dx} = \frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}} \).
This derivative is valid for \( x \in (-1, 1) \), where \( \sin^{-1}x \) is defined and differentiable.
In simple words: When a function is inside another function, like \( e \) raised to the power of inverse sine, we use the chain rule. We differentiate the outer function first, keeping the inner part as is, then multiply by the derivative of the inner function.

Exam Tip: The chain rule is essential for composite functions. Remember the standard derivatives like \( \frac{d}{dx}(e^x) = e^x \) and \( \frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}} \).

Question 3. Differentiate \( e^{x^{3}} \).
Answer: Let the function be \( y = e^{x^{3}} \). This is a composite function, so we will use the chain rule.
Let \( t = x^3 \).
Then the original function becomes \( y = e^t \).
Next, we find the derivatives of \( y \) with respect to \( t \) and \( t \) with respect to \( x \):
The derivative of \( y = e^t \) with respect to \( t \) is \( \frac{dy}{dt} = e^t \).
The derivative of \( t = x^3 \) with respect to \( x \) is \( \frac{dt}{dx} = 3x^2 \).
Now, apply the chain rule formula: \( \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} \).
Substitute the derivatives we calculated:
\( \frac{dy}{dx} = e^t \times 3x^2 \)
Finally, substitute \( t = x^3 \) back into the equation:

\( \implies \frac{dy}{dx} = 3x^2 e^{x^{3}} \).
In simple words: To find the derivative of \( e \) raised to a power like \( x^3 \), we differentiate \( e \) to that power (which stays the same) and then multiply it by the derivative of the power itself.

Exam Tip: When using the chain rule, clearly identify the "outer" function and the "inner" function. Differentiate the outer function with respect to the inner, then multiply by the derivative of the inner function with respect to \( x \).

Question 4. Find the derivative of \( \sin(\tan^{-1} e^{-x}) \).
Answer: Let the function be \( y = \sin(\tan^{-1} e^{-x}) \). This function involves three nested layers, so we apply the chain rule multiple times.
Let's define intermediate variables:
Let \( s = \tan^{-1} t \) and \( t = e^{-x} \).
Then the outermost function is \( y = \sin s \).
We find the derivative of each layer:
1. Derivative of \( y = \sin s \) with respect to \( s \): \( \frac{dy}{ds} = \cos s \).
2. Derivative of \( s = \tan^{-1} t \) with respect to \( t \): \( \frac{ds}{dt} = \frac{1}{1+t^2} \).
3. Derivative of \( t = e^{-x} \) with respect to \( x \): \( \frac{dt}{dx} = e^{-x} \cdot (-1) = -e^{-x} \).
Now, apply the triple chain rule: \( \frac{dy}{dx} = \frac{dy}{ds} \times \frac{ds}{dt} \times \frac{dt}{dx} \).
Substitute the derivatives:
\( \frac{dy}{dx} = (\cos s) \times \left(\frac{1}{1+t^2}\right) \times (-e^{-x}) \)
Next, substitute back the expressions for \( s \) and \( t \):
\( s = \tan^{-1} t = \tan^{-1}(e^{-x}) \)
\( t = e^{-x} \)

\( \implies \frac{dy}{dx} = \cos(\tan^{-1} e^{-x}) \times \frac{1}{1+(e^{-x})^2} \times (-e^{-x}) \)

\( \implies \frac{dy}{dx} = \frac{-e^{-x} \cos(\tan^{-1} e^{-x})}{1+e^{-2x}} \).
In simple words: For a function with many layers, like sine of inverse tangent of \( e \) to the power of \( -x \), we differentiate from the outside in. First sine, then inverse tangent, then \( e \) to the power of \( -x \). Each step gets multiplied by the derivative of its inner function.

Exam Tip: For deeply nested functions, clearly label your substitutions (\( y = \sin s \), \( s = \tan^{-1} t \), \( t = e^{-x} \)) to avoid errors in applying the chain rule sequentially.

Question 5. Differentiate \( \log(\cos(e^x)) \).
Answer: Let the function be \( y = \log(\cos(e^x)) \). This is another example requiring multiple applications of the chain rule.
Let's define intermediate variables for clarity:
Let \( s = \cos t \) and \( t = e^x \).
Then the outermost function is \( y = \log s \).
Now, we find the derivative of each component:
1. Derivative of \( y = \log s \) with respect to \( s \): \( \frac{dy}{ds} = \frac{1}{s} \).
2. Derivative of \( s = \cos t \) with respect to \( t \): \( \frac{ds}{dt} = -\sin t \).
3. Derivative of \( t = e^x \) with respect to \( x \): \( \frac{dt}{dx} = e^x \).
Apply the chain rule: \( \frac{dy}{dx} = \frac{dy}{ds} \times \frac{ds}{dt} \times \frac{dt}{dx} \).
Substitute the derivatives:
\( \frac{dy}{dx} = \left(\frac{1}{s}\right) \times (-\sin t) \times (e^x) \)
Next, substitute back the expressions for \( s \) and \( t \):
\( s = \cos t = \cos(e^x) \)
\( t = e^x \)

\( \implies \frac{dy}{dx} = \frac{1}{\cos(e^x)} \times (-\sin(e^x)) \times e^x \)

\( \implies \frac{dy}{dx} = \frac{-e^x \sin(e^x)}{\cos(e^x)} \)
Since \( \frac{\sin \theta}{\cos \theta} = \tan \theta \), we can simplify this further:

\( \implies \frac{dy}{dx} = -e^x \tan(e^x) \).
In simple words: To differentiate \( \log(\cos(e^x)) \), start from the outside: differentiate the log, then cosine, then \( e^x \). Multiply all these derivatives together, making sure to use the correct inner function at each step.

Exam Tip: Be mindful of the order of operations in differentiation and correctly identify the inner and outer functions. Double-check your basic derivative formulas for \( \log x \), \( \cos x \), and \( e^x \).

Question 6. Find the derivative of \( e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5} \).
Answer: Let the given function be \( y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5} \).
To find the derivative of a sum of functions, we differentiate each term separately and then add the results. That is, \( \frac{dy}{dx} = \frac{d}{dx}(e^x) + \frac{d}{dx}(e^{x^2}) + \frac{d}{dx}(e^{x^3}) + \frac{d}{dx}(e^{x^4}) + \frac{d}{dx}(e^{x^5}) \).
Let's find the derivative of a general term, \( e^{x^n} \), using the chain rule:
If \( u = e^{x^n} \), let \( t = x^n \). Then \( u = e^t \).
\( \frac{du}{dt} = e^t \) and \( \frac{dt}{dx} = nx^{n-1} \).
So, \( \frac{du}{dx} = \frac{du}{dt} \times \frac{dt}{dx} = e^t \times nx^{n-1} = nx^{n-1} e^{x^n} \).
Now, apply this rule to each term:
1. \( \frac{d}{dx}(e^x) = 1 \cdot x^{1-1} e^x = e^x \).
2. \( \frac{d}{dx}(e^{x^2}) = 2x^{2-1} e^{x^2} = 2x e^{x^2} \).
3. \( \frac{d}{dx}(e^{x^3}) = 3x^{3-1} e^{x^3} = 3x^2 e^{x^3} \).
4. \( \frac{d}{dx}(e^{x^4}) = 4x^{4-1} e^{x^4} = 4x^3 e^{x^4} \).
5. \( \frac{d}{dx}(e^{x^5}) = 5x^{5-1} e^{x^5} = 5x^4 e^{x^5} \).
Adding these derivatives together, we get:

\( \implies \frac{dy}{dx} = e^x + 2x e^{x^2} + 3x^2 e^{x^3} + 4x^3 e^{x^4} + 5x^4 e^{x^5} \).
In simple words: When you have many terms added together, you can find the derivative of each term separately and then sum up all those individual derivatives to get the total derivative. For each \( e \) raised to a power, you differentiate the exponent and multiply it by the original \( e \) expression.

Exam Tip: Remember the sum rule for derivatives: the derivative of a sum is the sum of the derivatives. For terms involving \( e^{f(x)} \), always apply the chain rule correctly, resulting in \( f'(x)e^{f(x)} \).

Question 7. Find the derivative of \( \sqrt{e^{\sqrt{x}}} \) where \( x > 0 \).
Answer: Let the function be \( y = \sqrt{e^{\sqrt{x}}} \). We can rewrite this as \( y = (e^{\sqrt{x}})^{\frac{1}{2}} \). This is a complex composite function, so we need to apply the chain rule multiple times.
Let's use a series of substitutions:
Let \( y = s^{\frac{1}{2}} \). Then \( s = e^t \). Then \( t = \sqrt{x} = x^{\frac{1}{2}} \).
Now, we find the derivative of each part:
1. Derivative of \( y = s^{\frac{1}{2}} \) with respect to \( s \): \( \frac{dy}{ds} = \frac{1}{2} s^{\frac{1}{2}-1} = \frac{1}{2} s^{-\frac{1}{2}} = \frac{1}{2\sqrt{s}} \).
2. Derivative of \( s = e^t \) with respect to \( t \): \( \frac{ds}{dt} = e^t \).
3. Derivative of \( t = x^{\frac{1}{2}} \) with respect to \( x \): \( \frac{dt}{dx} = \frac{1}{2} x^{\frac{1}{2}-1} = \frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}} \).
Apply the extended chain rule: \( \frac{dy}{dx} = \frac{dy}{ds} \times \frac{ds}{dt} \times \frac{dt}{dx} \).
Substitute the derivatives:
\( \frac{dy}{dx} = \left(\frac{1}{2\sqrt{s}}\right) \times (e^t) \times \left(\frac{1}{2\sqrt{x}}\right) \)
Now, substitute back the original expressions for \( s \) and \( t \):
\( s = e^t = e^{\sqrt{x}} \)
\( t = \sqrt{x} \)

\( \implies \frac{dy}{dx} = \frac{1}{2\sqrt{e^{\sqrt{x}}}} \times e^{\sqrt{x}} \times \frac{1}{2\sqrt{x}} \)
Combine the terms:
\( \frac{dy}{dx} = \frac{e^{\sqrt{x}}}{4\sqrt{x}\sqrt{e^{\sqrt{x}}}} \)
We know that \( \frac{A}{\sqrt{A}} = \sqrt{A} \), so \( \frac{e^{\sqrt{x}}}{\sqrt{e^{\sqrt{x}}}} = \sqrt{e^{\sqrt{x}}} \).

\( \implies \frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}} \).
This is valid for \( x > 0 \).
In simple words: This function has three layers: a square root, an exponential, and another square root. We differentiate each layer from the outside in, multiplying the results. For the outer square root, then the \( e \) function, and finally the inner square root.

Exam Tip: Rewrite square roots as powers of \( \frac{1}{2} \) to simplify differentiation using the power rule. When applying the chain rule multiple times, keep track of your substitutions to avoid errors.

Question 8. Differentiate \( \log(\log x) \) where \( x > 1 \).
Answer: Let the function be \( y = \log(\log x) \). This is a composite function, meaning one function is inside another, so we use the chain rule.
Let \( t = \log x \).
Then the expression becomes \( y = \log t \).
Now, we find the derivatives of \( y \) with respect to \( t \) and \( t \) with respect to \( x \):
The derivative of \( y = \log t \) with respect to \( t \) is \( \frac{dy}{dt} = \frac{1}{t} \).
The derivative of \( t = \log x \) with respect to \( x \) is \( \frac{dt}{dx} = \frac{1}{x} \).
Using the chain rule, \( \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} \).
Substitute the derivatives:
\( \frac{dy}{dx} = \frac{1}{t} \times \frac{1}{x} \)
Finally, substitute \( t = \log x \) back into the equation:

\( \implies \frac{dy}{dx} = \frac{1}{\log x} \times \frac{1}{x} \)

\( \implies \frac{dy}{dx} = \frac{1}{x \log x} \).
The condition \( x > 1 \) ensures that \( \log x > 0 \), so \( \log(\log x) \) is defined and its derivative exists.
In simple words: When you have a logarithm of another logarithm, you differentiate the outer log first (which means 1 divided by its inside part), and then multiply that by the derivative of the inner log function.

Exam Tip: For logarithmic functions, remember that \( \frac{d}{dx}(\log x) = \frac{1}{x} \). Pay attention to the domain conditions, especially for nested logarithms, to ensure the function is well-defined.

Question 9. Find the derivative of \( \frac{\cos x}{\log x} \) where \( x > 0 \).
Answer: Let the function be \( y = \frac{\cos x}{\log x} \). To differentiate this fractional form, we apply the quotient rule.
The quotient rule states that if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{u'v-uv'}{v^2} \).
Here, let \( u = \cos x \) and \( v = \log x \).
First, find the derivatives of \( u \) and \( v \):
The derivative of \( u \) with respect to \( x \) is \( u' = \frac{d}{dx}(\cos x) = -\sin x \).
The derivative of \( v \) with respect to \( x \) is \( v' = \frac{d}{dx}(\log x) = \frac{1}{x} \).
Now, substitute these into the quotient rule formula:
\( \frac{dy}{dx} = \frac{(-\sin x)(\log x) - (\cos x)(\frac{1}{x})}{(\log x)^2} \)
To simplify, multiply the numerator by \( x \) to remove the fraction in the second term:

\( \implies \frac{dy}{dx} = \frac{-x \sin x \log x - \cos x}{x(\log x)^2} \).
This derivative is valid for \( x > 0 \) and \( \log x \neq 0 \) (i.e., \( x \neq 1 \)).
In simple words: When differentiating a fraction, we use the quotient rule. We differentiate the top part, multiply by the bottom part, then subtract the top part multiplied by the derivative of the bottom part. All this is then divided by the square of the bottom part.

Exam Tip: Be careful with the signs when differentiating trigonometric functions, especially \( \frac{d}{dx}(\cos x) = -\sin x \). Ensure algebraic simplification of the numerator after applying the quotient rule, such as clearing fractions within the fraction.

Question 10. Differentiate \( \cos(\log x + e^x) \) where \( x > 0 \).
Answer: Let the function be \( y = \cos(\log x + e^x) \). This is a composite function, so we must use the chain rule.
Let \( t = \log x + e^x \).
Then the function becomes \( y = \cos t \).
Next, we find the derivatives of \( y \) with respect to \( t \) and \( t \) with respect to \( x \):
The derivative of \( y = \cos t \) with respect to \( t \) is \( \frac{dy}{dt} = -\sin t \).
The derivative of \( t = \log x + e^x \) with respect to \( x \) is found by differentiating each term:
\( \frac{dt}{dx} = \frac{d}{dx}(\log x) + \frac{d}{dx}(e^x) = \frac{1}{x} + e^x \).
Now, apply the chain rule formula: \( \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} \).
Substitute the derivatives we found:
\( \frac{dy}{dx} = (-\sin t) \times \left(\frac{1}{x} + e^x\right) \)
Finally, substitute \( t = \log x + e^x \) back into the equation:

\( \implies \frac{dy}{dx} = -\sin(\log x + e^x) \times \left(\frac{1}{x} + e^x\right) \)

\( \implies \frac{dy}{dx} = -\left(\frac{1}{x} + e^x\right) \sin(\log x + e^x) \).
This is valid for \( x > 0 \).
In simple words: To differentiate cosine of a sum of functions, first differentiate the cosine part (which becomes negative sine of the same sum), then multiply by the derivative of the entire sum inside the cosine. This inner derivative involves differentiating each term in the sum separately.

Exam Tip: When the inner function of a composite function is itself a sum, remember to differentiate each term of the sum individually before combining them as per the chain rule. Pay close attention to the negative sign when differentiating \( \cos t \).