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

Differentiate:

Question 1. \( x^2 \) w.r.t. \( x^3 \)
Answer: Let the first function be \( y = x^2 \) and the second function be \( z = x^3 \). We need to find the derivative of \( y \) with respect to \( z \), which means finding \( \frac{dy}{dz} \).
First, differentiate \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}(x^2) = 2x \)
Next, differentiate \( z \) with respect to \( x \):
\( \frac{dz}{dx} = \frac{d}{dx}(x^3) = 3x^2 \)
Now, use the chain rule to find \( \frac{dy}{dz} \):
\( \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \)
\( \implies \frac{dy}{dz} = \frac{2x}{3x^2} \)
\( \implies \frac{dy}{dz} = \frac{2}{3x} \) (assuming \( x \neq 0 \))
In simple words: We want to find how \( x^2 \) changes when \( x^3 \) changes. We first find how each changes with respect to \( x \), then divide the two results. This gives us the rate of change of \( x^2 \) with respect to \( x^3 \).

🎯 Exam Tip: Remember to express the final answer in terms of the variable of differentiation, or the original variable, as required. Always simplify algebraic fractions fully.

Question 2. \( x^3 - x^2 - x + 1 \) w.r.t. \( 3x^2 - x + 2 \)
Answer: Let the first function be \( y = x^3 - x^2 - x + 1 \) and the second function be \( z = 3x^2 - x + 2 \). We need to find \( \frac{dy}{dz} \).
First, differentiate \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}(x^3 - x^2 - x + 1) = 3x^2 - 2x - 1 \)
Next, differentiate \( z \) with respect to \( x \):
\( \frac{dz}{dx} = \frac{d}{dx}(3x^2 - x + 2) = 6x - 1 \)
Now, use the chain rule to find \( \frac{dy}{dz} \):
\( \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \)
\( \implies \frac{dy}{dz} = \frac{3x^2 - 2x - 1}{6x - 1} \) (This expression cannot be simplified further)
In simple words: To differentiate one function with respect to another, we find the individual rates of change for both functions using \( x \) and then divide them. This helps us see how one changes compared to the other.

🎯 Exam Tip: Be careful with the signs when differentiating polynomials, especially when terms are subtracted. Double-check each term's derivative before combining.

Question 3. \( \sin^3 x \) w.r.t. \( \cos^3 x \)
Answer: Let the first function be \( y = \sin^3 x \) and the second function be \( z = \cos^3 x \). We need to find \( \frac{dy}{dz} \).
First, differentiate \( y \) with respect to \( x \) using the chain rule (\( (u^n)' = n u^{n-1} u' \)):
\( \frac{dy}{dx} = 3 \sin^2 x \cdot \frac{d}{dx}(\sin x) = 3 \sin^2 x \cos x \)
Next, differentiate \( z \) with respect to \( x \) using the chain rule:
\( \frac{dz}{dx} = 3 \cos^2 x \cdot \frac{d}{dx}(\cos x) = 3 \cos^2 x (-\sin x) \)
Now, use the chain rule to find \( \frac{dy}{dz} \):
\( \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \)
\( \implies \frac{dy}{dz} = \frac{3 \sin^2 x \cos x}{3 \cos^2 x (-\sin x)} \)
Cancel out common terms (assuming \( \sin x \neq 0 \) and \( \cos x \neq 0 \)):
\( \implies \frac{dy}{dz} = \frac{\sin x}{-\cos x} \)
\( \implies \frac{dy}{dz} = -\tan x \)
In simple words: We are finding how \( \sin^3 x \) changes as \( \cos^3 x \) changes. After differentiating both parts separately and dividing, we simplify the expression using trigonometry to get the final answer.

🎯 Exam Tip: When using the chain rule for trigonometric powers, remember to differentiate both the power and the trigonometric function itself. Simplification using identities like \( \frac{\sin x}{\cos x} = \tan x \) is crucial.

Question 4. \( \sin x^3 \) w.r.t. \( \sec^2 x^2 \)
Answer: Let the first function be \( y = \sin(x^3) \) and the second function be \( z = \sec^2(x^2) \). We need to find \( \frac{dy}{dz} \).
First, differentiate \( y \) with respect to \( x \) using the chain rule:
\( \frac{dy}{dx} = \cos(x^3) \cdot \frac{d}{dx}(x^3) = \cos(x^3) \cdot (3x^2) = 3x^2 \cos(x^3) \)
Next, differentiate \( z \) with respect to \( x \) using the chain rule (for \( (u^n)' \) and then \( \sec(v)' \)):
\( \frac{dz}{dx} = 2 \sec(x^2) \cdot \frac{d}{dx}(\sec(x^2)) \)
\( \implies \frac{dz}{dx} = 2 \sec(x^2) \cdot (\sec(x^2) \tan(x^2)) \cdot \frac{d}{dx}(x^2) \)
\( \implies \frac{dz}{dx} = 2 \sec^2(x^2) \tan(x^2) \cdot (2x) = 4x \sec^2(x^2) \tan(x^2) \)
Now, use the chain rule to find \( \frac{dy}{dz} \):
\( \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \)
\( \implies \frac{dy}{dz} = \frac{3x^2 \cos(x^3)}{4x \sec^2(x^2) \tan(x^2)} \)
Simplify the expression (assuming \( x \neq 0 \)):
\( \implies \frac{dy}{dz} = \frac{3x \cos(x^3)}{4 \sec^2(x^2) \tan(x^2)} \)
In simple words: This problem involves nested functions, so we use the chain rule multiple times for each part. We break down the differentiation into smaller steps for \( y \) and \( z \) with respect to \( x \), and then combine them to get the final answer.

🎯 Exam Tip: When differentiating complex functions like \( \sec^2(x^2) \), apply the chain rule layer by layer: first the power, then the trigonometric function, then its argument. Don't forget any part of the chain.

Question 5. Differentiate
(i) \( \tan^{-1}\left(\frac{2 x}{1-x^2}\right) \) w.r.t. \( \tan^{-1} x \)
(ii) \( \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \) w.r.t. \( \sec^{-1}\left(\frac{1}{2 x^2-1}\right) \)
(iii) \( \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) w.r.t. \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) \)
(iv) \( \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \) w.r.t. \( \tan^{-1} x \)
Answer:
(i) Let \( y = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) and \( z = \tan^{-1} x \).
To simplify \( y \), substitute \( x = \tan \theta \), which means \( \theta = \tan^{-1} x \).
Then \( y = \tan^{-1}\left(\frac{2 \tan \theta}{1-\tan^2 \theta}\right) \)
This is the formula for \( \tan(2\theta) \):
\( \implies y = \tan^{-1}(\tan 2\theta) \)
\( \implies y = 2\theta \)
Substitute back \( \theta = \tan^{-1} x \):
\( \implies y = 2 \tan^{-1} x \)
Now, differentiate \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}(2 \tan^{-1} x) = \frac{2}{1+x^2} \)
Next, differentiate \( z \) with respect to \( x \):
\( \frac{dz}{dx} = \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2} \)
Finally, find \( \frac{dy}{dz} \):
\( \frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{2/(1+x^2)}{1/(1+x^2)} \)
\( \implies \frac{dy}{dz} = 2 \)

(ii) Let \( y = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \) and \( z = \sec^{-1}\left(\frac{1}{2x^2-1}\right) \).
To simplify \( y \), substitute \( x = \sin \theta \), so \( \theta = \sin^{-1} x \).
\( y = \tan^{-1}\left(\frac{\sin \theta}{\sqrt{1-\sin^2 \theta}}\right) = \tan^{-1}\left(\frac{\sin \theta}{\sqrt{\cos^2 \theta}}\right) = \tan^{-1}\left(\frac{\sin \theta}{\cos \theta}\right) \)
\( \implies y = \tan^{-1}(\tan \theta) = \theta \)
Substitute back \( \theta = \sin^{-1} x \):
\( \implies y = \sin^{-1} x \)
Differentiate \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}} \)
To simplify \( z \), substitute \( x = \cos \phi \), so \( \phi = \cos^{-1} x \).
\( z = \sec^{-1}\left(\frac{1}{2\cos^2 \phi - 1}\right) \)
This is the formula for \( \cos(2\phi) \):
\( \implies z = \sec^{-1}\left(\frac{1}{\cos 2\phi}\right) = \sec^{-1}(\sec 2\phi) \)
\( \implies z = 2\phi \)
Substitute back \( \phi = \cos^{-1} x \):
\( \implies z = 2 \cos^{-1} x \)
Differentiate \( z \) with respect to \( x \):
\( \frac{dz}{dx} = \frac{d}{dx}(2 \cos^{-1} x) = 2 \cdot \frac{-1}{\sqrt{1-x^2}} = \frac{-2}{\sqrt{1-x^2}} \)
Finally, find \( \frac{dy}{dz} \):
\( \frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{1/\sqrt{1-x^2}}{-2/\sqrt{1-x^2}} \)
\( \implies \frac{dy}{dz} = -\frac{1}{2} \)

(iii) Let \( y = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) and \( z = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \).
To simplify \( y \), substitute \( x = \tan \theta \), so \( \theta = \tan^{-1} x \).
\( y = \tan^{-1}\left(\frac{2 \tan \theta}{1-\tan^2 \theta}\right) = \tan^{-1}(\tan 2\theta) = 2\theta \)
\( \implies y = 2 \tan^{-1} x \)
Differentiate \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}(2 \tan^{-1} x) = \frac{2}{1+x^2} \)
To simplify \( z \), substitute \( x = \tan \phi \), so \( \phi = \tan^{-1} x \).
\( z = \sin^{-1}\left(\frac{2 \tan \phi}{1+\tan^2 \phi}\right) \)
This is the formula for \( \sin(2\phi) \):
\( \implies z = \sin^{-1}(\sin 2\phi) = 2\phi \)
Substitute back \( \phi = \tan^{-1} x \):
\( \implies z = 2 \tan^{-1} x \)
Differentiate \( z \) with respect to \( x \):
\( \frac{dz}{dx} = \frac{d}{dx}(2 \tan^{-1} x) = \frac{2}{1+x^2} \)
Finally, find \( \frac{dy}{dz} \):
\( \frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{2/(1+x^2)}{2/(1+x^2)} \)
\( \implies \frac{dy}{dz} = 1 \)

(iv) Let \( y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \) and \( z = \tan^{-1} x \).
To simplify \( y \), substitute \( x = \tan \theta \), so \( \theta = \tan^{-1} x \).
\( y = \tan^{-1}\left(\frac{\sqrt{1+\tan^2 \theta}-1}{\tan \theta}\right) \)
Using the identity \( 1+\tan^2 \theta = \sec^2 \theta \):
\( \implies y = \tan^{-1}\left(\frac{\sqrt{\sec^2 \theta}-1}{\tan \theta}\right) = \tan^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \)
Convert to sine and cosine:
\( \implies y = \tan^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right) = \tan^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \)
Using half-angle formulas: \( 1-\cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right) \) and \( \sin \theta = 2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right) \)
\( \implies y = \tan^{-1}\left(\frac{2 \sin^2(\theta/2)}{2 \sin(\theta/2) \cos(\theta/2)}\right) \)
\( \implies y = \tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) = \frac{\theta}{2} \)
Substitute back \( \theta = \tan^{-1} x \):
\( \implies y = \frac{1}{2} \tan^{-1} x \)
Differentiate \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2} \tan^{-1} x\right) = \frac{1}{2} \cdot \frac{1}{1+x^2} = \frac{1}{2(1+x^2)} \)
Differentiate \( z \) with respect to \( x \):
\( \frac{dz}{dx} = \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2} \)
Finally, find \( \frac{dy}{dz} \):
\( \frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{1/(2(1+x^2))}{1/(1+x^2)} \)
\( \implies \frac{dy}{dz} = \frac{1}{2} \)
In simple words: For all these questions, we use trigonometric substitutions to simplify the inverse trigonometric functions. After simplifying, the differentiation becomes much easier. The key is to recognize the correct trigonometric identity that matches the expression.

🎯 Exam Tip: Mastering inverse trigonometric substitution formulas (e.g., \( x=\tan\theta \) for \( \tan^{-1}\frac{2x}{1-x^2} \)) is crucial for these types of differentiation problems. Remember to convert back to \( x \) after simplifying with \( \theta \) or \( \phi \).