Samacheer Kalvi Class 11 Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.3

Differentiate the following:

Question 1. Differentiate \( y = (x^2 + 4x + 6)^5 \).
Answer: Let \( u = x^2 + 4x + 6 \). We first find the derivative of \( u \) with respect to \( x \).
\( \frac{du}{dx} = 2x + 4 \)
Now, we have \( y = u^5 \). We find the derivative of \( y \) with respect to \( u \).
\( \frac{dy}{du} = 5u^4 \)
Using the chain rule, which helps find the derivative of composite functions:
\( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)
\( \implies \frac{dy}{dx} = 5u^4 (2x + 4) \)
Substitute back \( u = x^2 + 4x + 6 \):
\( \implies \frac{dy}{dx} = 5(x^2 + 4x + 6)^4 (2x + 4) \)
We can rearrange the terms for a clearer final form.
\( \implies \frac{dy}{dx} = 5(2x + 4)(x^2 + 4x + 6)^4 \)
In simple words: To differentiate this function, we use the chain rule. We first differentiate the inner part, then the outer part, and multiply the results together.

🎯 Exam Tip: Remember to apply the chain rule correctly when differentiating composite functions. Differentiate from the outside in, then multiply by the derivative of the inner function.

Question 2. Differentiate \( y = \tan 3x \).
Answer: This is a composite function, so we will use the chain rule. Let \( u = 3x \).
Then \( \frac{du}{dx} = 3 \).
Now, \( y = \tan u \). So, \( \frac{dy}{du} = \sec^2 u \).
Using the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
\( \implies \frac{dy}{dx} = \sec^2 (3x) \times 3 \)
\( \implies \frac{dy}{dx} = 3 \sec^2 3x \)
This rule helps us differentiate functions that are "functions of functions".
In simple words: We find the derivative of tan (something) which is sec squared (something). Then we multiply it by the derivative of the "something" itself.

🎯 Exam Tip: When differentiating \( \tan(ax+b) \), the result will always be \( a \sec^2(ax+b) \). Recognize this pattern for quick solutions.

Question 3. Differentiate \( y = \cos (\tan x) \).
Answer: We use the chain rule for this composite function. Let \( u = \tan x \).
Then, we find the derivative of \( u \) with respect to \( x \):
\( \frac{du}{dx} = \sec^2 x \)
Now, \( y = \cos u \). We find the derivative of \( y \) with respect to \( u \):
\( \frac{dy}{du} = -\sin u \)
Applying the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
\( \implies \frac{dy}{dx} = (-\sin u) (\sec^2 x) \)
Substitute back \( u = \tan x \):
\( \implies \frac{dy}{dx} = -\sin(\tan x) \cdot \sec^2 x \)
The order of multiplication does not change the result, but writing the trigonometric function first is common.
In simple words: For a cosine of a function, we take the derivative of cosine (which is minus sine), keep the function inside, and then multiply by the derivative of that inner function.

🎯 Exam Tip: Be careful with signs. The derivative of \( \cos x \) is \( -\sin x \), which is a common mistake point.

Question 4. Differentiate \( y = \sqrt[3]{1+x^{3}} \).
Answer: First, rewrite the function using fractional exponents: \( y = (1+x^3)^{\frac{1}{3}} \).
We use the chain rule. Let \( u = 1+x^3 \).
Then, we find the derivative of \( u \) with respect to \( x \):
\( \frac{du}{dx} = 3x^2 \)
Now, \( y = u^{\frac{1}{3}} \). We find the derivative of \( y \) with respect to \( u \):
\( \frac{dy}{du} = \frac{1}{3} u^{\frac{1}{3}-1} = \frac{1}{3} u^{-\frac{2}{3}} \)
Applying the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
\( \implies \frac{dy}{dx} = \frac{1}{3} (1+x^3)^{-\frac{2}{3}} \cdot (3x^2) \)
\( \implies \frac{dy}{dx} = x^2 (1+x^3)^{-\frac{2}{3}} \)
This can also be written with a positive exponent in the denominator, showing another common way to present the answer.
\( \implies \frac{dy}{dx} = \frac{x^2}{(1+x^3)^{\frac{2}{3}}} \)
In simple words: To differentiate a cube root of a function, we treat it like that function raised to the power of one-third. We use the chain rule, differentiating the outer power first, then multiplying by the derivative of the inner part.

🎯 Exam Tip: When dealing with roots, always convert them to fractional exponents first (e.g., \( \sqrt[n]{x} = x^{\frac{1}{n}} \)) as it makes differentiation using the power rule much easier.

Question 5. Differentiate \( y = e^{\sqrt{x}} \).
Answer: This is a composite function, so we will use the chain rule. First, rewrite \( \sqrt{x} \) as \( x^{\frac{1}{2}} \). So, \( y = e^{x^{\frac{1}{2}}} \).
Let \( u = \sqrt{x} = x^{\frac{1}{2}} \).
Then, we find the derivative of \( u \) with respect to \( x \):
\( \frac{du}{dx} = \frac{1}{2} x^{\frac{1}{2}-1} = \frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}} \)
Now, \( y = e^u \). We find the derivative of \( y \) with respect to \( u \):
\( \frac{dy}{du} = e^u \)
Applying the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
\( \implies \frac{dy}{dx} = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} \)
\( \implies \frac{dy}{dx} = \frac{e^{\sqrt{x}}}{2\sqrt{x}} \)
This demonstrates how the chain rule applies to exponential functions with complex powers.
In simple words: For \( e \) raised to a power, its derivative is \( e \) to that same power, multiplied by the derivative of the power itself.

🎯 Exam Tip: Remember that the derivative of \( e^x \) is \( e^x \). When the exponent is a function of \( x \) (like \( \sqrt{x} \)), always apply the chain rule by multiplying by the derivative of that exponent.

Question 6. Differentiate \( y = \sin (e^x) \).
Answer: We use the chain rule for this composite function. Let \( u = e^x \).
Then, we find the derivative of \( u \) with respect to \( x \):
\( \frac{du}{dx} = e^x \)
Now, \( y = \sin u \). We find the derivative of \( y \) with respect to \( u \):
\( \frac{dy}{du} = \cos u \)
Applying the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
\( \implies \frac{dy}{dx} = \cos (e^x) \cdot e^x \)
\( \implies \frac{dy}{dx} = e^x \cos (e^x) \)
The chain rule helps differentiate functions where one function is "inside" another, like \( \sin \) of an exponential function.
In simple words: For sine of a function, we take the derivative of sine (which is cosine), keep the function inside, and then multiply by the derivative of that inner function.

🎯 Exam Tip: Be careful not to confuse \( \sin(e^x) \) with \( e^{\sin x} \) or \( (\sin x)^e \). Each requires a different application of differentiation rules.

Question 7. Differentiate \( F(x) = (x^3 + 4x)^7 \).
Answer: We use the chain rule. Let \( u = x^3 + 4x \).
Then, we find the derivative of \( u \) with respect to \( x \):
\( \frac{du}{dx} = 3x^2 + 4 \)
Now, \( F(x) = u^7 \). We find the derivative of \( F(x) \) with respect to \( u \):
\( \frac{dF}{du} = 7u^{7-1} = 7u^6 \)
Applying the chain rule, \( \frac{dF}{dx} = \frac{dF}{du} \times \frac{du}{dx} \).
\( \implies F'(x) = 7(x^3 + 4x)^6 \cdot (3x^2 + 4) \)
This method is especially useful for powers of functions, where the "power rule" combines with the chain rule.
In simple words: For a function raised to a power, we first apply the power rule (bring the power down, reduce the power by one) to the whole bracket, and then multiply by the derivative of what's inside the bracket.

🎯 Exam Tip: Always remember to multiply by the derivative of the inner function when using the chain rule, especially with power functions. A common mistake is to forget this final multiplication.

Question 8. Differentiate \( h (t) = \left(t-\frac{1}{t}\right)^{\frac{3}{2}} \).
Answer: Rewrite the function as \( h(t) = (t - t^{-1})^{\frac{3}{2}} \). We will use the chain rule.
Let \( u = t - \frac{1}{t} = t - t^{-1} \).
Then, we find the derivative of \( u \) with respect to \( t \):
\( \frac{du}{dt} = 1 - (-1)t^{-1-1} = 1 + t^{-2} = 1 + \frac{1}{t^2} \)
Now, \( h(t) = u^{\frac{3}{2}} \). We find the derivative of \( h(t) \) with respect to \( u \):
\( \frac{dh}{du} = \frac{3}{2} u^{\frac{3}{2}-1} = \frac{3}{2} u^{\frac{1}{2}} \)
Applying the chain rule, \( \frac{dh}{dt} = \frac{dh}{du} \times \frac{du}{dt} \).
\( \implies h'(t) = \frac{3}{2} \left(t-\frac{1}{t}\right)^{\frac{1}{2}} \cdot \left(1 + \frac{1}{t^2}\right) \)
This problem combines the power rule with the chain rule and negative exponents.
In simple words: First, rewrite \( 1/t \) as \( t \) to the power of minus one. Then, use the chain rule: differentiate the outer power function, and multiply by the derivative of the inner function.

🎯 Exam Tip: When differentiating terms like \( \frac{1}{t} \), convert them to \( t^{-1} \) to apply the power rule easily. Remember that \( \frac{d}{dt}(t^{-1}) = -t^{-2} \).

Question 9. Differentiate \( f(t) = \sqrt[3]{1+\tan t} \).
Answer: First, rewrite the function using fractional exponents: \( f(t) = (1+\tan t)^{\frac{1}{3}} \).
We use the chain rule. Let \( u = 1+\tan t \).
Then, we find the derivative of \( u \) with respect to \( t \):
\( \frac{du}{dt} = 0 + \sec^2 t = \sec^2 t \)
Now, \( f(t) = u^{\frac{1}{3}} \). We find the derivative of \( f(t) \) with respect to \( u \):
\( \frac{df}{du} = \frac{1}{3} u^{\frac{1}{3}-1} = \frac{1}{3} u^{-\frac{2}{3}} \)
Applying the chain rule, \( \frac{df}{dt} = \frac{df}{du} \times \frac{du}{dt} \).
\( \implies f'(t) = \frac{1}{3} (1+\tan t)^{-\frac{2}{3}} \cdot (\sec^2 t) \)
\( \implies f'(t) = \frac{\sec^2 t}{3(1+\tan t)^{\frac{2}{3}}} \)
The final answer is usually given with positive exponents, which often means moving terms to the denominator.
In simple words: Convert the cube root into a power. Then use the chain rule, first differentiating the power and then multiplying by the derivative of the inside function.

🎯 Exam Tip: Always convert roots to fractional exponents before differentiating. Remember the derivative of \( \tan x \) is \( \sec^2 x \), which is a key trigonometric derivative.

Question 10. Differentiate \( y = \cos (a^3 + x^3) \).
Answer: We use the chain rule for this composite function. Let \( u = a^3 + x^3 \). Note that \( a \) is a constant.
Then, we find the derivative of \( u \) with respect to \( x \):
\( \frac{du}{dx} = 0 + 3x^2 = 3x^2 \)
Now, \( y = \cos u \). We find the derivative of \( y \) with respect to \( u \):
\( \frac{dy}{du} = -\sin u \)
Applying the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
\( \implies \frac{dy}{dx} = -\sin(a^3 + x^3) \cdot (3x^2) \)
\( \implies \frac{dy}{dx} = -3x^2 \sin(a^3 + x^3) \)
Understanding the role of constants in differentiation is crucial for accurate results.
In simple words: We take the derivative of cosine (which is negative sine), keeping the bracketed part the same. Then, we multiply this by the derivative of the content inside the bracket.

🎯 Exam Tip: Remember that the derivative of a constant term (like \( a^3 \)) is always zero. Only terms involving \( x \) are differentiated with respect to \( x \).

Question 11. Differentiate \( y = e^{-mx} \).
Answer: We use the chain rule for this composite function. Let \( u = -mx \).
Then, we find the derivative of \( u \) with respect to \( x \):
\( \frac{du}{dx} = -m \)
Now, \( y = e^u \). We find the derivative of \( y \) with respect to \( u \):
\( \frac{dy}{du} = e^u \)
Applying the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
\( \implies \frac{dy}{dx} = e^{-mx} \cdot (-m) \)
\( \implies \frac{dy}{dx} = -m e^{-mx} \)
Since \( y = e^{-mx} \), we can also write the derivative in terms of \( y \).
\( \implies \frac{dy}{dx} = -my \)
In simple words: The derivative of \( e \) to the power of "something" is \( e \) to that same power, multiplied by the derivative of the "something" itself.

🎯 Exam Tip: The derivative of \( e^{ax} \) is \( ae^{ax} \). This is a very common result in calculus, so it's good to memorize it.

Question 12. Differentiate \( y = 4 \sec 5x \).
Answer: We use the chain rule, keeping the constant multiplier \( 4 \) at the front. Let \( u = 5x \).
Then, we find the derivative of \( u \) with respect to \( x \):
\( \frac{du}{dx} = 5 \)
Now, \( y = 4 \sec u \). We find the derivative of \( y \) with respect to \( u \):
\( \frac{dy}{du} = 4 \sec u \tan u \)
Applying the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
\( \implies \frac{dy}{dx} = (4 \sec 5x \tan 5x) \cdot 5 \)
\( \implies \frac{dy}{dx} = 20 \sec 5x \tan 5x \)
This shows how constants are handled in differentiation and the application of the chain rule to trigonometric functions.
In simple words: We take the derivative of secant (which is secant times tangent), keep the inner function the same, and then multiply by the derivative of that inner function. Don't forget to multiply by the constant out front.

🎯 Exam Tip: Remember the derivative of \( \sec x \) is \( \sec x \tan x \). When there's a constant multiplier like \( 4 \), it simply stays multiplied throughout the differentiation process.

Question 13. Differentiate \( y = (2x – 5)^4 (8x^2 – 5)^{-3} \).
Answer: We will use the product rule, which states that if \( y = uv \), then \( \frac{dy}{dx} = u'v + uv' \).
Let \( u = (2x - 5)^4 \) and \( v = (8x^2 - 5)^{-3} \).
First, find the derivatives of \( u \) and \( v \) using the chain rule:
\( u' = \frac{d}{dx} (2x - 5)^4 = 4(2x - 5)^3 \cdot (2) = 8(2x - 5)^3 \)
\( v' = \frac{d}{dx} (8x^2 - 5)^{-3} = -3(8x^2 - 5)^{-4} \cdot (16x) = -48x(8x^2 - 5)^{-4} \)
Now, apply the product rule:
\( \frac{dy}{dx} = u'v + uv' \)
\( \implies \frac{dy}{dx} = 8(2x - 5)^3 (8x^2 - 5)^{-3} + (2x - 5)^4 (-48x(8x^2 - 5)^{-4}) \)
\( \implies \frac{dy}{dx} = 8(2x - 5)^3 (8x^2 - 5)^{-3} - 48x(2x - 5)^4 (8x^2 - 5)^{-4} \)
Factor out the common terms, which are \( 8(2x - 5)^3 (8x^2 - 5)^{-4} \):
\( \implies \frac{dy}{dx} = 8(2x - 5)^3 (8x^2 - 5)^{-4} \left[ (8x^2 - 5)^1 - 6x(2x - 5)^1 \right] \)
Simplify the expression inside the square brackets:
\( (8x^2 - 5) - 6x(2x - 5) = 8x^2 - 5 - 12x^2 + 30x = -4x^2 + 30x - 5 \)
Substitute this back into the derivative:
\( \implies \frac{dy}{dx} = 8(2x - 5)^3 (8x^2 - 5)^{-4} (-4x^2 + 30x - 5) \)
Rewrite with positive exponents:
\( \implies \frac{dy}{dx} = \frac{8(2x - 5)^3 (-4x^2 + 30x - 5)}{(8x^2 - 5)^4} \)
Combining the rules for products and powers is a common task in advanced differentiation.
In simple words: When differentiating a product of two functions, we use the product rule. This involves finding the derivative of each function separately, then combining them as (derivative of first * second function) + (first function * derivative of second). Don't forget the chain rule for each part.

🎯 Exam Tip: For product rule problems, identify \( u \) and \( v \) clearly, then find their derivatives \( u' \) and \( v' \) first. This breaks the problem into smaller, manageable steps, reducing errors.

Question 14. Differentiate \( y = (x^2 + 1) \sqrt[3]{x^{2}+2} \).
Answer: First, rewrite the function as \( y = (x^2 + 1) (x^2 + 2)^{\frac{1}{3}} \).
We will use the product rule \( \frac{dy}{dx} = u'v + uv' \).
Let \( u = x^2 + 1 \) and \( v = (x^2 + 2)^{\frac{1}{3}} \).
Find the derivatives of \( u \) and \( v \):
\( u' = \frac{d}{dx} (x^2 + 1) = 2x \)
\( v' = \frac{d}{dx} (x^2 + 2)^{\frac{1}{3}} = \frac{1}{3}(x^2 + 2)^{-\frac{2}{3}} \cdot (2x) = \frac{2x}{3(x^2 + 2)^{\frac{2}{3}}} \)
Now apply the product rule:
\( \frac{dy}{dx} = (2x)(x^2 + 2)^{\frac{1}{3}} + (x^2 + 1) \left( \frac{2x}{3(x^2 + 2)^{\frac{2}{3}}} \right) \)
To combine these terms, find a common denominator, which is \( 3(x^2 + 2)^{\frac{2}{3}} \).
\( \frac{dy}{dx} = \frac{2x(x^2 + 2)^{\frac{1}{3}} \cdot 3(x^2 + 2)^{\frac{2}{3}}}{3(x^2 + 2)^{\frac{2}{3}}} + \frac{2x(x^2 + 1)}{3(x^2 + 2)^{\frac{2}{3}}} \)
When multiplying powers with the same base, add the exponents: \( (x^2+2)^{\frac{1}{3}} \cdot (x^2+2)^{\frac{2}{3}} = (x^2+2)^{\frac{1}{3}+\frac{2}{3}} = (x^2+2)^1 \).
\( \implies \frac{dy}{dx} = \frac{6x(x^2 + 2) + 2x(x^2 + 1)}{3(x^2 + 2)^{\frac{2}{3}}} \)
Expand and simplify the numerator:
\( 6x(x^2 + 2) + 2x(x^2 + 1) = 6x^3 + 12x + 2x^3 + 2x = 8x^3 + 14x \)
So, the final derivative is:
\( \implies \frac{dy}{dx} = \frac{8x^3 + 14x}{3(x^2 + 2)^{\frac{2}{3}}} \)
This problem tests both the product rule and the chain rule along with algebraic simplification.
In simple words: First, change the cube root into a power. Then, use the product rule for differentiating. Make sure to apply the chain rule when finding the derivative of the second part. Combine everything by finding a common denominator.

🎯 Exam Tip: When combining terms with fractional exponents, remember to find a common denominator. Also, be careful with exponent addition: \( x^a \cdot x^b = x^{a+b} \).

Question 15. Differentiate \( y = x e^{-x^2} \).
Answer: We will use the product rule \( \frac{dy}{dx} = u'v + uv' \).
Let \( u = x \) and \( v = e^{-x^2} \).
Find the derivatives of \( u \) and \( v \):
\( u' = \frac{d}{dx} (x) = 1 \)
\( v' = \frac{d}{dx} (e^{-x^2}) \). For this, use the chain rule. Let \( w = -x^2 \). Then \( \frac{dw}{dx} = -2x \).
So, \( \frac{dv}{dx} = e^w \cdot \frac{dw}{dx} = e^{-x^2} \cdot (-2x) = -2xe^{-x^2} \).
Now apply the product rule:
\( \frac{dy}{dx} = (1) (e^{-x^2}) + (x) (-2xe^{-x^2}) \)
\( \implies \frac{dy}{dx} = e^{-x^2} - 2x^2 e^{-x^2} \)
Factor out the common term \( e^{-x^2} \):
\( \implies \frac{dy}{dx} = e^{-x^2} (1 - 2x^2) \)
This example effectively combines the product rule with the chain rule for exponential functions.
In simple words: This is a product of two functions, so we use the product rule. For the exponential part, remember to use the chain rule because its power is also a function of \( x \).

🎯 Exam Tip: Exponential functions like \( e^{f(x)} \) are always differentiated using the chain rule, resulting in \( f'(x)e^{f(x)} \). This often appears within other rules like the product rule.

Question 16. Differentiate \( s(t) = \sqrt[4]{\frac{t^{3}+1}{t^{3}-1}} \).
Answer: First, rewrite the function using fractional exponents: \( s(t) = \left(\frac{t^3+1}{t^3-1}\right)^{\frac{1}{4}} \).
We will use the chain rule and the quotient rule. Let \( u = \frac{t^3+1}{t^3-1} \).
Then \( s(t) = u^{\frac{1}{4}} \). The derivative of \( s(t) \) with respect to \( u \) is:
\( \frac{ds}{du} = \frac{1}{4} u^{\frac{1}{4}-1} = \frac{1}{4} u^{-\frac{3}{4}} \)
Now, find the derivative of \( u \) with respect to \( t \) using the quotient rule: \( \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \).
Let \( f = t^3+1 \) and \( g = t^3-1 \). Then \( f' = 3t^2 \) and \( g' = 3t^2 \).
\( \frac{du}{dt} = \frac{(3t^2)(t^3-1) - (t^3+1)(3t^2)}{(t^3-1)^2} \)
\( \implies \frac{du}{dt} = \frac{3t^5 - 3t^2 - (3t^5 + 3t^2)}{(t^3-1)^2} \)
\( \implies \frac{du}{dt} = \frac{3t^5 - 3t^2 - 3t^5 - 3t^2}{(t^3-1)^2} \)
\( \implies \frac{du}{dt} = \frac{-6t^2}{(t^3-1)^2} \)
Now, apply the chain rule \( \frac{ds}{dt} = \frac{ds}{du} \times \frac{du}{dt} \):
\( s'(t) = \frac{1}{4} \left(\frac{t^3+1}{t^3-1}\right)^{-\frac{3}{4}} \cdot \left(\frac{-6t^2}{(t^3-1)^2}\right) \)
Rewrite the negative exponent and simplify:
\( s'(t) = \frac{1}{4} \frac{(t^3-1)^{\frac{3}{4}}}{(t^3+1)^{\frac{3}{4}}} \cdot \frac{-6t^2}{(t^3-1)^2} \)
\( \implies s'(t) = \frac{-6t^2}{4 (t^3+1)^{\frac{3}{4}} (t^3-1)^{2 - \frac{3}{4}}} \)
\( \implies s'(t) = \frac{-6t^2}{4 (t^3+1)^{\frac{3}{4}} (t^3-1)^{\frac{5}{4}}} \)
\( \implies s'(t) = \frac{-3t^2}{2 (t^3+1)^{\frac{3}{4}} (t^3-1)^{\frac{5}{4}}} \)
This problem involves both the chain rule and the quotient rule, requiring careful application of both.
In simple words: First, convert the fourth root to a power. Then, use the chain rule for the overall function. The derivative of the inner part needs the quotient rule, as it's a fraction of two functions. Multiply these two derivatives together for the final answer.

🎯 Exam Tip: For functions involving roots of quotients, it's often easiest to convert the root to a fractional exponent first. Then, apply the chain rule for the outer power and the quotient rule for the inner fraction.

Question 17. Differentiate \( f(x) = \frac{x}{\sqrt{7-3 x}} \).
Answer: First, rewrite the function to avoid the quotient rule and use the product rule instead: \( f(x) = x (7-3x)^{-\frac{1}{2}} \).
We will use the product rule \( f'(x) = u'v + uv' \).
Let \( u = x \) and \( v = (7-3x)^{-\frac{1}{2}} \).
Find the derivatives of \( u \) and \( v \):
\( u' = \frac{d}{dx} (x) = 1 \)
\( v' = \frac{d}{dx} (7-3x)^{-\frac{1}{2}} \). Use the chain rule for this part.
\( v' = -\frac{1}{2}(7-3x)^{-\frac{1}{2}-1} \cdot (-3) = -\frac{1}{2}(7-3x)^{-\frac{3}{2}} \cdot (-3) = \frac{3}{2}(7-3x)^{-\frac{3}{2}} \)
Now apply the product rule:
\( f'(x) = (1)(7-3x)^{-\frac{1}{2}} + (x)\left(\frac{3}{2}(7-3x)^{-\frac{3}{2}}\right) \)
\( \implies f'(x) = (7-3x)^{-\frac{1}{2}} + \frac{3x}{2}(7-3x)^{-\frac{3}{2}} \)
Factor out the common term with the lowest power, which is \( (7-3x)^{-\frac{3}{2}} \):
\( f'(x) = (7-3x)^{-\frac{3}{2}} \left[ (7-3x)^1 + \frac{3x}{2} \right] \)
Simplify the expression inside the square brackets:
\( (7-3x) + \frac{3x}{2} = 7 - 3x + 1.5x = 7 - 1.5x = 7 - \frac{3x}{2} \)
To combine, get a common denominator in the bracket: \( \frac{14}{2} - \frac{3x}{2} = \frac{14 - 3x}{2} \).
Substitute this back:
\( \implies f'(x) = (7-3x)^{-\frac{3}{2}} \left( \frac{14 - 3x}{2} \right) \)
Rewrite with positive exponents:
\( \implies f'(x) = \frac{14 - 3x}{2(7-3x)^{\frac{3}{2}}} \)
This solution demonstrates how converting to fractional exponents and using the product rule can simplify differentiation.
In simple words: Change the square root in the bottom to a negative power. Then, use the product rule because it's like two functions multiplied together. Make sure to apply the chain rule when differentiating the part with the power. Simplify by finding a common factor.

🎯 Exam Tip: Often, converting a quotient with a root in the denominator into a product with a negative fractional exponent simplifies the differentiation process, allowing the use of the product rule and chain rule more easily than the quotient rule.

Question 18. Differentiate \( y = \tan (\cos x) \).
Answer: We use the chain rule for this composite function. Let \( u = \cos x \).
Then, we find the derivative of \( u \) with respect to \( x \):
\( \frac{du}{dx} = -\sin x \)
Now, \( y = \tan u \). We find the derivative of \( y \) with respect to \( u \):
\( \frac{dy}{du} = \sec^2 u \)
Applying the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
\( \implies \frac{dy}{dx} = \sec^2 (\cos x) \cdot (-\sin x) \)
\( \implies \frac{dy}{dx} = -\sin x \sec^2 (\cos x) \)
The chain rule is essential for finding derivatives of functions where one function is embedded within another.
In simple words: We find the derivative of tangent (which is secant squared), keeping the inner part the same. Then, we multiply this by the derivative of the inner part.

🎯 Exam Tip: Be careful with the signs in trigonometric derivatives; the derivative of \( \cos x \) is \( -\sin x \), which is a common source of error if missed.

Question 19. Differentiate \( y = \frac{\sin^2 x}{\cos x} \).
Answer: We will use the quotient rule \( \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \).
Let \( f = \sin^2 x \) and \( g = \cos x \).
Find the derivatives of \( f \) and \( g \):
\( f' = \frac{d}{dx} (\sin x)^2 = 2(\sin x)^1 \cdot (\cos x) = 2 \sin x \cos x \)
\( g' = \frac{d}{dx} (\cos x) = -\sin x \)
Now apply the quotient rule:
\( \frac{dy}{dx} = \frac{(2 \sin x \cos x)(\cos x) - (\sin^2 x)(-\sin x)}{(\cos x)^2} \)
\( \implies \frac{dy}{dx} = \frac{2 \sin x \cos^2 x + \sin^3 x}{\cos^2 x} \)
Factor out \( \sin x \) from the numerator:
\( \implies \frac{dy}{dx} = \frac{\sin x (2 \cos^2 x + \sin^2 x)}{\cos^2 x} \)
Separate the terms in the numerator to simplify:
\( \implies \frac{dy}{dx} = \sin x \left( \frac{2 \cos^2 x}{\cos^2 x} + \frac{\sin^2 x}{\cos^2 x} \right) \)
\( \implies \frac{dy}{dx} = \sin x (2 + \tan^2 x) \)
Using the identity \( 1 + \tan^2 x = \sec^2 x \), we can further simplify the expression in the parenthesis:
\( 2 + \tan^2 x = 1 + (1 + \tan^2 x) = 1 + \sec^2 x \).
So, the derivative is:
\( \implies \frac{dy}{dx} = \sin x (1 + \sec^2 x) \)
This problem demonstrates using the quotient rule, chain rule for \( \sin^2 x \), and trigonometric identities for simplification.
In simple words: Since it's a fraction, use the quotient rule. Remember to use the chain rule for \( \sin^2 x \). After applying the rules, simplify the expression using trigonometric identities like \( \sin^2 x + \cos^2 x = 1 \) and \( 1 + \tan^2 x = \sec^2 x \).

🎯 Exam Tip: Always be on the lookout for trigonometric identities to simplify your answer. A derivative might look complex initially, but identities can often lead to a much cleaner final form.

Question 20. Differentiate \( y = 5^{\frac{-1}{x}} \).
Answer: First, rewrite the function in terms of \( e \) using the identity \( a^b = e^{b \ln a} \):
\( y = e^{\ln(5^{-\frac{1}{x}})} = e^{-\frac{1}{x} \ln 5} \)
Now, we use the chain rule. Let \( u = -\frac{1}{x} \ln 5 = -x^{-1} \ln 5 \).
Find the derivative of \( u \) with respect to \( x \):
\( \frac{du}{dx} = -(-1)x^{-1-1} \ln 5 = x^{-2} \ln 5 = \frac{\ln 5}{x^2} \)
Now, \( y = e^u \). We find the derivative of \( y \) with respect to \( u \):
\( \frac{dy}{du} = e^u \)
Applying the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
\( \implies \frac{dy}{dx} = e^{-\frac{1}{x} \ln 5} \cdot \frac{\ln 5}{x^2} \)
Substitute back \( e^{-\frac{1}{x} \ln 5} = 5^{-\frac{1}{x}} \):
\( \implies \frac{dy}{dx} = 5^{-\frac{1}{x}} \cdot \frac{\ln 5}{x^2} \)
This type of problem shows how to differentiate exponential functions with bases other than \( e \).
In simple words: Rewrite the function using \( e \) and natural logarithm. Then, use the chain rule. The derivative of \( e \) to a power is \( e \) to that power, multiplied by the derivative of the power itself. Remember to convert back to the original base if needed.

🎯 Exam Tip: When differentiating \( a^{f(x)} \), rewrite it as \( e^{f(x) \ln a} \) and then apply the chain rule. The result will be \( a^{f(x)} \cdot f'(x) \cdot \ln a \).

Question 21. Differentiate \( y = \sqrt{1+2 \tan x} \).
Answer: First, rewrite the function using fractional exponents: \( y = (1+2 \tan x)^{\frac{1}{2}} \).
We use the chain rule. Let \( u = 1+2 \tan x \).
Then, we find the derivative of \( u \) with respect to \( x \):
\( \frac{du}{dx} = 0 + 2 \sec^2 x = 2 \sec^2 x \)
Now, \( y = u^{\frac{1}{2}} \). We find the derivative of \( y \) with respect to \( u \):
\( \frac{dy}{du} = \frac{1}{2} u^{\frac{1}{2}-1} = \frac{1}{2} u^{-\frac{1}{2}} \)
Applying the chain rule, \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
\( \implies \frac{dy}{dx} = \frac{1}{2} (1+2 \tan x)^{-\frac{1}{2}} \cdot (2 \sec^2 x) \)
\( \implies \frac{dy}{dx} = \frac{2 \sec^2 x}{2 (1+2 \tan x)^{\frac{1}{2}}} \)
\( \implies \frac{dy}{dx} = \frac{\sec^2 x}{\sqrt{1+2 \tan x}} \)
This problem shows how the chain rule applies when a trigonometric function is part of an outer root function.
In simple words: Change the square root to a power of one-half. Use the chain rule by differentiating the outer power first, then multiplying by the derivative of the inside part. Remember that the derivative of \( \tan x \) is \( \sec^2 x \).

🎯 Exam Tip: Always remember that the derivative of \( \tan x \) is \( \sec^2 x \). This is a common derivative used in conjunction with the chain rule for square root functions.

Question 22. Differentiate \( y = \sin^3 x + \cos^3 x \).
Answer: We differentiate each term separately. For each term, we use the chain rule (power rule followed by the derivative of the base function).
Let \( y = u + v \), where \( u = \sin^3 x = (\sin x)^3 \) and \( v = \cos^3 x = (\cos x)^3 \).
First, find \( \frac{du}{dx} \):
\( \frac{du}{dx} = 3(\sin x)^2 \cdot (\cos x) = 3 \sin^2 x \cos x \)
Next, find \( \frac{dv}{dx} \):
\( \frac{dv}{dx} = 3(\cos x)^2 \cdot (-\sin x) = -3 \sin x \cos^2 x \)
Now, sum the derivatives:
\( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} = 3 \sin^2 x \cos x - 3 \sin x \cos^2 x \)
Factor out the common terms, which are \( 3 \sin x \cos x \):
\( \implies \frac{dy}{dx} = 3 \sin x \cos x (\sin x - \cos x) \)
This problem demonstrates differentiating a sum of power-of-trigonometric functions.
In simple words: Differentiate each part separately. For \( \sin^3 x \), treat it as \( (\sin x)^3 \). Use the power rule first, then multiply by the derivative of \( \sin x \). Do the same for \( \cos^3 x \), remembering the derivative of \( \cos x \) is negative sine. Then, combine and factor.

🎯 Exam Tip: When differentiating \( \sin^n x \) or \( \cos^n x \), remember it's a chain rule application: \( n \sin^{n-1} x \cdot \cos x \) and \( n \cos^{n-1} x \cdot (-\sin x) \) respectively. Factoring common terms at the end can simplify the answer.

Question 23. Differentiate \( y = \sin^2 (\cos kx) \).
Answer: This is a triple composite function. We apply the chain rule multiple times.
Let \( y = [\sin(\cos kx)]^2 \).
**Step 1:** Differentiate the outermost power function, \( (\cdot)^2 \).
\( \frac{dy}{d(\sin(\cos kx))} = 2 \sin(\cos kx) \)
**Step 2:** Differentiate the \( \sin(\cdot) \) function.
\( \frac{d(\sin(\cos kx))}{d(\cos kx)} = \cos(\cos kx) \)
**Step 3:** Differentiate the \( \cos(\cdot) \) function.
\( \frac{d(\cos kx)}{d(kx)} = -\sin kx \)
**Step 4:** Differentiate the innermost function, \( kx \).
\( \frac{d(kx)}{dx} = k \)
Multiply all these derivatives together:
\( \frac{dy}{dx} = [2 \sin(\cos kx)] \cdot [\cos(\cos kx)] \cdot [-\sin kx] \cdot [k] \)
Rearrange the terms:
\( \implies \frac{dy}{dx} = -2k \sin(\cos kx) \cos(\cos kx) \sin kx \)
We can use the double angle identity \( \sin(2A) = 2 \sin A \cos A \). Here, \( A = \cos kx \).
So, \( 2 \sin(\cos kx) \cos(\cos kx) = \sin(2 \cos kx) \).
\( \implies \frac{dy}{dx} = -k \sin kx \sin(2 \cos kx) \)
This problem demonstrates a complex application of the chain rule and trigonometric identities.
In simple words: This function has three layers. Start from the outside: differentiate the square power, then the sine function, then the cosine function, and finally the \( kx \) part. Multiply all these derivatives. You can also use a double angle identity to simplify the answer.

🎯 Exam Tip: For nested functions like \( f(g(h(x))) \), apply the chain rule layer by layer: \( f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \). Be systematic and remember trigonometric identities to simplify.

Question 24. Differentiate \( y = (1 + \cos 2x)^6 \).
Answer: We use the chain rule. Let \( u = 1 + \cos 2x \).
Then \( y = u^6 \). The derivative of \( y \) with respect to \( u \) is:
\( \frac{dy}{du} = 6u^{6-1} = 6u^5 \)
Now, find the derivative of \( u \) with respect to \( x \): \( \frac{du}{dx} = \frac{d}{dx} (1 + \cos 2x) \).
The derivative of \( 1 \) is \( 0 \). For \( \cos 2x \), use the chain rule again: \( \frac{d}{dx}(\cos 2x) = -\sin 2x \cdot 2 = -2 \sin 2x \).
So, \( \frac{du}{dx} = 0 - 2 \sin 2x = -2 \sin 2x \).
Apply the chain rule \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \):
\( \implies \frac{dy}{dx} = 6(1 + \cos 2x)^5 \cdot (-2 \sin 2x) \)
\( \implies \frac{dy}{dx} = -12 \sin 2x (1 + \cos 2x)^5 \)
This problem shows how the chain rule can be nested, differentiating the outermost function first, then the inner function.
In simple words: This function is a power of another function. First, apply the power rule (bring the power down, reduce it by one) to the whole bracket. Then, multiply by the derivative of what's inside the bracket, remembering to use the chain rule again for the cosine part.

🎯 Exam Tip: When differentiating \( (f(x))^n \), the chain rule gives \( n(f(x))^{n-1} \cdot f'(x) \). Ensure you correctly find \( f'(x) \) which might also require the chain rule itself.

Question 25. Differentiate \( y = \frac{e^{3 x}}{1+e^{x}} \).
Answer: We will use the quotient rule \( \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \).
Let \( f = e^{3x} \) and \( g = 1+e^x \).
Find the derivatives of \( f \) and \( g \):
\( f' = \frac{d}{dx} (e^{3x}) = 3e^{3x} \)
\( g' = \frac{d}{dx} (1+e^x) = 0 + e^x = e^x \)
Now apply the quotient rule:
\( \frac{dy}{dx} = \frac{(3e^{3x})(1+e^x) - (e^{3x})(e^x)}{(1+e^x)^2} \)
Expand the numerator:
\( \implies \frac{dy}{dx} = \frac{3e^{3x} + 3e^{3x}e^x - e^{3x}e^x}{(1+e^x)^2} \)
Combine the terms in the numerator (remember \( e^a e^b = e^{a+b} \)):
\( \implies \frac{dy}{dx} = \frac{3e^{3x} + 3e^{4x} - e^{4x}}{(1+e^x)^2} \)
\( \implies \frac{dy}{dx} = \frac{3e^{3x} + 2e^{4x}}{(1+e^x)^2} \)
We can factor out \( e^{3x} \) from the numerator for a more concise form:
\( \implies \frac{dy}{dx} = \frac{e^{3x} (3 + 2e^x)}{(1+e^x)^2} \)
This problem demonstrates the application of the quotient rule to exponential functions.
In simple words: Since it's a fraction, use the quotient rule. Find the derivative of the top and bottom parts separately. Then combine them using the quotient rule formula and simplify the expression.

🎯 Exam Tip: Remember the basic derivatives: \( \frac{d}{dx} e^{ax} = ae^{ax} \) and \( \frac{d}{dx} (constant) = 0 \). These are fundamental when applying the quotient rule to exponential functions.

Question 26. Differentiate the following: \( y = \sqrt{x+\sqrt{x}} \)
Answer:
We have the function \( y = \sqrt{x+\sqrt{x}} \).
This can be rewritten using fractional exponents as \( y = \left(x+x^{\frac{1}{2}}\right)^{\frac{1}{2}} \).
We will use the chain rule, which states that if \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
Let \( u = x+\sqrt{x} \), so \( y = \sqrt{u} = u^{\frac{1}{2}} \).
First, find \( \frac{dy}{du} \):
\( \frac{dy}{du} = \frac{1}{2} u^{\frac{1}{2}-1} = \frac{1}{2} u^{-\frac{1}{2}} = \frac{1}{2\sqrt{u}} \).
Next, find \( \frac{du}{dx} \):
\( \frac{du}{dx} = \frac{d}{dx} (x+x^{\frac{1}{2}}) = 1 + \frac{1}{2} x^{\frac{1}{2}-1} = 1 + \frac{1}{2} x^{-\frac{1}{2}} = 1 + \frac{1}{2\sqrt{x}} \).
Now, apply the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
\( \frac{dy}{dx} = \frac{1}{2\sqrt{x+\sqrt{x}}} \times \left(1 + \frac{1}{2\sqrt{x}}\right) \)
Combine the terms:
\( \frac{dy}{dx} = \frac{1}{2\sqrt{x+\sqrt{x}}} \times \left(\frac{2\sqrt{x}+1}{2\sqrt{x}}\right) \)
\( \frac{dy}{dx} = \frac{2\sqrt{x}+1}{4\sqrt{x}\sqrt{x+\sqrt{x}}} \).
This is the derivative of the given function. It is important to correctly apply the chain rule multiple times for nested functions.
In simple words: To find the derivative of this function, we break it down. First, we treat the inner part as a single unit and find the derivative of the outer square root. Then, we find the derivative of that inner part itself. Finally, we multiply these two results together to get the full answer.

🎯 Exam Tip: When differentiating nested functions like \( \sqrt{f(x)} \), remember to use the chain rule: \( \frac{d}{dx} \sqrt{f(x)} = \frac{1}{2\sqrt{f(x)}} \cdot f'(x) \). Always simplify the inner derivative first.

Question 27. Differentiate the following: \( y = e^x \cos x \)
Answer:
We need to differentiate the function \( y = e^x \cos x \).
This function is a product of two functions, \( u = e^x \) and \( v = \cos x \).
We will use the product rule for differentiation, which states: \( \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \).
First, find the derivative of \( u = e^x \):
\( \frac{du}{dx} = \frac{d}{dx} (e^x) = e^x \).
Next, find the derivative of \( v = \cos x \):
\( \frac{dv}{dx} = \frac{d}{dx} (\cos x) = -\sin x \).
Now, apply the product rule:
\( \frac{dy}{dx} = e^x (-\sin x) + (\cos x) (e^x) \)
\( \frac{dy}{dx} = -e^x \sin x + e^x \cos x \)
We can factor out \( e^x \) from both terms:
\( \frac{dy}{dx} = e^x (\cos x - \sin x) \).
This shows the derivative of the product of two functions.
In simple words: To differentiate this, we use the product rule. This means we take the first function and multiply it by the derivative of the second, then add the second function multiplied by the derivative of the first. In this case, it gives us \( e^x \) multiplied by (cosine minus sine).

🎯 Exam Tip: The product rule is crucial for functions multiplied together. Remember the formula \( (uv)' = u'v + uv' \) and carefully apply it to each part.

Question 28. Differentiate the following: \( y = \sqrt{x+\sqrt{x+\sqrt{x}}} \)
Answer:
We are asked to differentiate the function \( y = \sqrt{x+\sqrt{x+\sqrt{x}}} \).
This is a complex nested function, requiring multiple applications of the chain rule.
Let's rewrite the function using fractional exponents:
\( y = \left[x+\left(x+x^{\frac{1}{2}}\right)^{\frac{1}{2}}\right]^{\frac{1}{2}} \).
We use the chain rule \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
Let \( u = x+\sqrt{x+\sqrt{x}} \), then \( y = \sqrt{u} = u^{\frac{1}{2}} \).
So, \( \frac{dy}{dx} = \frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}} \cdot \frac{d}{dx}(x+\sqrt{x+\sqrt{x}}) \).
Now we need to find \( \frac{d}{dx}(x+\sqrt{x+\sqrt{x}}) \):
\( \frac{d}{dx}(x+\sqrt{x+\sqrt{x}}) = 1 + \frac{d}{dx}(\sqrt{x+\sqrt{x}}) \).
Next, we find \( \frac{d}{dx}(\sqrt{x+\sqrt{x}}) \). Let \( v = x+\sqrt{x} \), then this is \( \sqrt{v} = v^{\frac{1}{2}} \).
\( \frac{d}{dx}(\sqrt{x+\sqrt{x}}) = \frac{1}{2\sqrt{x+\sqrt{x}}} \cdot \frac{d}{dx}(x+\sqrt{x}) \).
Then, find \( \frac{d}{dx}(x+\sqrt{x}) \):
\( \frac{d}{dx}(x+\sqrt{x}) = 1 + \frac{d}{dx}(x^{\frac{1}{2}}) = 1 + \frac{1}{2}x^{-\frac{1}{2}} = 1 + \frac{1}{2\sqrt{x}} \).
Now, substitute back step-by-step:
\( \frac{d}{dx}(\sqrt{x+\sqrt{x}}) = \frac{1}{2\sqrt{x+\sqrt{x}}} \cdot \left(1 + \frac{1}{2\sqrt{x}}\right) \)
\( \frac{d}{dx}(\sqrt{x+\sqrt{x}}) = \frac{1}{2\sqrt{x+\sqrt{x}}} \cdot \left(\frac{2\sqrt{x}+1}{2\sqrt{x}}\right) = \frac{2\sqrt{x}+1}{4\sqrt{x}\sqrt{x+\sqrt{x}}} \).
Now, substitute this back into the derivative of the second term:
\( \frac{d}{dx}(x+\sqrt{x+\sqrt{x}}) = 1 + \frac{2\sqrt{x}+1}{4\sqrt{x}\sqrt{x+\sqrt{x}}} \).
Finally, substitute everything back into the main derivative for \( y \):
\( \frac{dy}{dx} = \frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}} \cdot \left(1 + \frac{2\sqrt{x}+1}{4\sqrt{x}\sqrt{x+\sqrt{x}}}\right) \).
This can be combined into a single fraction:
\( \frac{dy}{dx} = \frac{1}{2\sqrt{x+\sqrt{x+\sqrt{x}}}} \cdot \left(\frac{4\sqrt{x}\sqrt{x+\sqrt{x}} + 2\sqrt{x}+1}{4\sqrt{x}\sqrt{x+\sqrt{x}}}\right) \)
\( \implies \frac{dy}{dx} = \frac{4\sqrt{x}\sqrt{x+\sqrt{x}} + 2\sqrt{x}+1}{8\sqrt{x}\sqrt{x+\sqrt{x}}\sqrt{x+\sqrt{x+\sqrt{x}}}} \).
This is the derivative, showing how repeated application of the chain rule is necessary for deeply nested functions.
In simple words: This problem involves a square root inside a square root, which is inside another square root. We solve it by peeling the layers one by one, starting from the outermost square root and working our way inwards. For each layer, we use the chain rule, which means we differentiate the outer part and then multiply by the derivative of what's inside.

🎯 Exam Tip: For deeply nested functions, always identify the "outer" and "inner" functions carefully. Apply the chain rule step-by-step from outside to inside, making sure not to miss any layers of differentiation.

Question 29. Differentiate the following: \( y = \sin (\tan (\sqrt{\sin x})) \)
Answer:
We need to find the derivative of \( y = \sin (\tan (\sqrt{\sin x})) \).
This function is a composition of several functions, requiring repeated use of the chain rule.
The chain rule states: \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
Here, the outermost function is sine, then tangent, then square root, then sine.
Step 1: Differentiate the outermost function, \( \sin(u) \), where \( u = \tan (\sqrt{\sin x}) \).
\( \frac{d}{du}(\sin u) = \cos u \).
So, the first part is \( \cos (\tan (\sqrt{\sin x})) \).
Step 2: Differentiate the next inner function, \( \tan(v) \), where \( v = \sqrt{\sin x} \).
\( \frac{d}{dv}(\tan v) = \sec^2 v \).
So, the second part is \( \sec^2 (\sqrt{\sin x}) \).
Step 3: Differentiate the next inner function, \( \sqrt{w} \), where \( w = \sin x \).
\( \frac{d}{dw}(\sqrt{w}) = \frac{1}{2\sqrt{w}} \).
So, the third part is \( \frac{1}{2\sqrt{\sin x}} \).
Step 4: Differentiate the innermost function, \( \sin x \).
\( \frac{d}{dx}(\sin x) = \cos x \).
Now, multiply all these derivatives together according to the chain rule:
\( \frac{dy}{dx} = \cos (\tan (\sqrt{\sin x})) \times \sec^2 (\sqrt{\sin x}) \times \frac{1}{2\sqrt{\sin x}} \times \cos x \).
Rearranging the terms for clarity:
\( \frac{dy}{dx} = \frac{\cos x \cdot \cos (\tan (\sqrt{\sin x})) \cdot \sec^2 (\sqrt{\sin x})}{2\sqrt{\sin x}} \).
This shows the complete derivative by applying the chain rule multiple times for nested functions.
In simple words: To find the derivative of this very nested function, we go step-by-step from the outside to the inside. We first differentiate the 'sin' part, then the 'tan' part, then the square root part, and finally the inner 'sin x' part. We multiply all these smaller derivatives together to get the final answer.

🎯 Exam Tip: When faced with multiple nested functions, systematically differentiate from the outermost layer inward. Write down each partial derivative and then multiply them all together to complete the chain rule application.

Question 30. Differentiate the following: \( y = \sin^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \)
Answer:
We need to differentiate the function \( y = \sin^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \).
This involves the derivative of \( \sin^{-1}(u) \) and the quotient rule for \( u = \frac{1-x^2}{1+x^2} \).
The derivative of \( \sin^{-1}(u) \) is \( \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \).
Let \( u = \frac{1-x^2}{1+x^2} \).
First, find \( \frac{du}{dx} \) using the quotient rule: \( \left(\frac{P}{Q}\right)' = \frac{P'Q - PQ'}{Q^2} \).
Here, \( P = 1-x^2 \) and \( Q = 1+x^2 \).
\( P' = \frac{d}{dx}(1-x^2) = -2x \).
\( Q' = \frac{d}{dx}(1+x^2) = 2x \).
\( \frac{du}{dx} = \frac{(-2x)(1+x^2) - (1-x^2)(2x)}{(1+x^2)^2} \)
\( \implies \frac{du}{dx} = \frac{-2x - 2x^3 - (2x - 2x^3)}{(1+x^2)^2} \)
\( \implies \frac{du}{dx} = \frac{-2x - 2x^3 - 2x + 2x^3}{(1+x^2)^2} \)
\( \implies \frac{du}{dx} = \frac{-4x}{(1+x^2)^2} \).
Now, calculate \( \sqrt{1-u^2} \):
\( \sqrt{1-u^2} = \sqrt{1 - \left(\frac{1-x^2}{1+x^2}\right)^2} \)
\( \implies \sqrt{1 - \frac{(1-x^2)^2}{(1+x^2)^2}} = \sqrt{\frac{(1+x^2)^2 - (1-x^2)^2}{(1+x^2)^2}} \)
\( \implies \sqrt{\frac{(1+2x^2+x^4) - (1-2x^2+x^4)}{(1+x^2)^2}} = \sqrt{\frac{1+2x^2+x^4 - 1+2x^2-x^4}{(1+x^2)^2}} \)
\( \implies \sqrt{\frac{4x^2}{(1+x^2)^2}} = \frac{\sqrt{4x^2}}{\sqrt{(1+x^2)^2}} = \frac{2|x|}{1+x^2} \).
Since usually \( x \) is considered positive in these contexts, we can write \( \frac{2x}{1+x^2} \).
Finally, combine everything to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{1}{\frac{2x}{1+x^2}} \times \frac{-4x}{(1+x^2)^2} \)
\( \implies \frac{dy}{dx} = \frac{1+x^2}{2x} \times \frac{-4x}{(1+x^2)^2} \)
Cancel common terms:
\( \implies \frac{dy}{dx} = \frac{-4x}{2x(1+x^2)} = \frac{-2}{1+x^2} \).
This shows the derivative of the inverse sine function with a rational expression inside.
In simple words: This problem asks us to differentiate an inverse sine function. First, we find the derivative of the fraction inside the inverse sine using the quotient rule. Then, we use the formula for differentiating inverse sine, which involves a square root. We combine these parts and simplify the whole expression to get the final answer.

🎯 Exam Tip: Recognize the standard derivative of \( \sin^{-1}(u) \) and apply the chain rule. The expression \( \frac{1-x^2}{1+x^2} \) is often involved in substitution problems (e.g., \( x=\tan\theta \)), which can simplify the derivative calculation.