RBSE Solutions Class 12 Maths Chapter 7 Differentiation Exercise 7.2

Question 1. Find derivative of following functions w.r.t. x :
(i) \( \sin^{-1} \{2x\sqrt{1-x^2}\}, \frac{1}{2} < x < \frac{1}{2} \)
(ii) \( \sin^{-1} (3x-4x^3), x\in(1) \)
Answer:
(i) Let \( y = \sin^{-1} (2x\sqrt{1-x^2}) \).
We substitute \( x = \sin\theta \). This means \( \theta = \sin^{-1}x \).
So, \( y = \sin^{-1} (2\sin\theta\sqrt{1-\sin^2\theta}) \)
\( y = \sin^{-1} (2\sin\theta\cos\theta) \) (since \( \sqrt{1-\sin^2\theta} = \sqrt{\cos^2\theta} = \cos\theta \))
\( y = \sin^{-1} (\sin 2\theta) \)
\( y = 2\theta \)
Now, we substitute back \( \theta = \sin^{-1}x \).
\( y = 2\sin^{-1}x \)
Next, we differentiate \( y \) with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}(2\sin^{-1}x) \)
\( \frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{1-x^2}} \)
\( \frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}} \)
(ii) Let \( y = \sin^{-1} (3x-4x^3) \).
We substitute \( x = \sin\theta \). This means \( \theta = \sin^{-1}x \).
So, \( y = \sin^{-1} (3\sin\theta - 4\sin^3\theta) \)
We know the trigonometric identity \( \sin 3\theta = 3\sin\theta - 4\sin^3\theta \).
\( y = \sin^{-1} (\sin 3\theta) \)
\( y = 3\theta \)
Now, we substitute back \( \theta = \sin^{-1}x \).
\( y = 3\sin^{-1}x \)
Next, we differentiate \( y \) with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}(3\sin^{-1}x) \)
\( \frac{dy}{dx} = 3 \cdot \frac{1}{\sqrt{1-x^2}} \)
\( \frac{dy}{dx} = \frac{3}{\sqrt{1-x^2}} \)
In simple words: For both parts, we use a trigonometric substitution to simplify the expression inside the inverse sine function. Once simplified, we differentiate the resulting simple inverse trigonometric function using standard rules.

🎯 Exam Tip: Remember common trigonometric identities like \( \sin 2\theta = 2\sin\theta\cos\theta \) and \( \sin 3\theta = 3\sin\theta - 4\sin^3\theta \) for quick simplification in inverse trigonometric differentiation problems.

Question 2. Find derivative of following functions w.r.t. x :
(i) \( \cos^{-1}\left(\frac{2x}{1-x^2}\right), x \in (-1, 1) \)
(ii) \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), x \in (0,1) \)
Answer:
(i) Let \( y = \cos^{-1}\left(\frac{2x}{1-x^2}\right) \).
We substitute \( x = \tan\theta \). This means \( \theta = \tan^{-1}x \).
So, \( y = \cos^{-1}\left(\frac{2\tan\theta}{1-\tan^2\theta}\right) \)
We know the trigonometric identity \( \tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta} \). So, \( y = \cos^{-1}(\tan 2\theta) \).
This expression cannot be directly simplified as \( \tan 2\theta \) is not in the form of cosine. We must use another identity.
We know \( \tan A = \cot\left(\frac{\pi}{2}-A\right) \) and \( \cot A = \tan\left(\frac{\pi}{2}-A\right) \).
Also, \( \cos^{-1} (\sin A) = \cos^{-1} (\cos(\frac{\pi}{2}-A)) = \frac{\pi}{2}-A \).
So, \( y = \cos^{-1}(\sin 2\theta) \)
\( y = \cos^{-1}\left\{\cos\left(\frac{\pi}{2}-2\theta\right)\right\} \)
\( y = \frac{\pi}{2}-2\theta \)
Now, we substitute back \( \theta = \tan^{-1}x \).
\( y = \frac{\pi}{2}-2\tan^{-1}x \)
Next, we differentiate \( y \) with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}-2\tan^{-1}x\right) \)
\( \frac{dy}{dx} = 0 - 2 \cdot \frac{1}{1+x^2} \)
\( \frac{dy}{dx} = -\frac{2}{1+x^2} \)
(ii) Let \( y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \).
We substitute \( x = \tan\theta \). This means \( \theta = \tan^{-1}x \).
So, \( y = \cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) \)
We know the trigonometric identity \( \cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta} \).
\( y = \cos^{-1}(\cos 2\theta) \)
\( y = 2\theta \)
Now, we substitute back \( \theta = \tan^{-1}x \).
\( y = 2\tan^{-1}x \)
Next, we differentiate \( y \) with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}(2\tan^{-1}x) \)
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} \)
\( \frac{dy}{dx} = \frac{2}{1+x^2} \)
In simple words: For these problems, we use the substitution \( x = \tan\theta \) to turn the complex inverse cosine expressions into simpler trigonometric forms like \( \sin 2\theta \) or \( \cos 2\theta \). Then, we convert these back to inverse tangent functions and differentiate.

🎯 Exam Tip: Recognizing the double angle formulas for tangent, sine, and cosine in terms of tangent of the half-angle is key to simplifying these expressions. For example, \( \frac{2\tan\theta}{1-\tan^2\theta} = \tan 2\theta \) and \( \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta \).

Question 3. Find the derivative of the following functions with respect to x:
(i) \( 3\cos^{-1}x \)
(ii) \( \cos^{-1}\sqrt{\frac{1+x}{2}} \)
Answer:
(i) Let \( y = 3\cos^{-1}x \).
To find the derivative, we use the standard differentiation formula for \( \cos^{-1}x \), which is \( \frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}} \).
\( \frac{dy}{dx} = \frac{d}{dx}(3\cos^{-1}x) \)
\( \frac{dy}{dx} = 3 \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) \)
\( \frac{dy}{dx} = -\frac{3}{\sqrt{1-x^2}} \)
(ii) Let \( y = \cos^{-1}\sqrt{\frac{1+x}{2}} \).
We substitute \( x = \cos\theta \). This means \( \theta = \cos^{-1}x \).
So, \( y = \cos^{-1}\sqrt{\frac{1+\cos\theta}{2}} \)
We know the half-angle trigonometric identity \( \cos^2\left(\frac{\theta}{2}\right) = \frac{1+\cos\theta}{2} \).
\( y = \cos^{-1}\sqrt{\cos^2\left(\frac{\theta}{2}\right)} \)
\( y = \cos^{-1}\left(\cos\left(\frac{\theta}{2}\right)\right) \)
\( y = \frac{\theta}{2} \)
Now, we substitute back \( \theta = \cos^{-1}x \).
\( y = \frac{1}{2}\cos^{-1}x \)
Next, we differentiate \( y \) with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2}\cos^{-1}x\right) \)
\( \frac{dy}{dx} = \frac{1}{2} \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) \)
\( \frac{dy}{dx} = -\frac{1}{2\sqrt{1-x^2}} \)
In simple words: For the first part, we directly apply the derivative formula for inverse cosine. For the second part, we use a substitution involving cosine and a half-angle identity to simplify the expression before differentiating.

🎯 Exam Tip: Recognizing expressions like \( \sqrt{\frac{1+\cos\theta}{2}} \) as half-angle formulas is crucial. These simplifications make differentiation much easier than using the chain rule directly.

Question 4. Find the derivative of the following functions with respect to x:
(i) \( \sec^{-1}\left(\frac{1}{2x^2-1}\right) \)
(ii) \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \)
Answer:
(i) Let \( y = \sec^{-1}\left(\frac{1}{2x^2-1}\right) \).
We substitute \( x = \cos\theta \). This means \( \theta = \cos^{-1}x \).
So, \( y = \sec^{-1}\left(\frac{1}{2\cos^2\theta-1}\right) \)
We know the trigonometric identity \( \cos 2\theta = 2\cos^2\theta-1 \).
\( y = \sec^{-1}\left(\frac{1}{\cos 2\theta}\right) \)
We know that \( \frac{1}{\cos A} = \sec A \).
\( y = \sec^{-1}(\sec 2\theta) \)
\( y = 2\theta \)
Now, we substitute back \( \theta = \cos^{-1}x \).
\( y = 2\cos^{-1}x \)
Next, we differentiate \( y \) with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}(2\cos^{-1}x) \)
\( \frac{dy}{dx} = 2 \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right) \)
\( \frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}} \)
(ii) Let \( y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \).
We substitute \( x = \tan\theta \). This means \( \theta = \tan^{-1}x \).
So, \( y = \cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) \)
We know the trigonometric identity \( \cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta} \).
\( y = \cos^{-1}(\cos 2\theta) \)
\( y = 2\theta \)
Now, we substitute back \( \theta = \tan^{-1}x \).
\( y = 2\tan^{-1}x \)
Next, we differentiate \( y \) with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}(2\tan^{-1}x) \)
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} \)
\( \frac{dy}{dx} = \frac{2}{1+x^2} \)
In simple words: For these problems, we use trigonometric substitutions like \( x = \cos\theta \) or \( x = \tan\theta \) to transform the expressions into simpler forms involving double angle identities. This allows us to reduce the inverse trigonometric functions to linear forms which are then easily differentiated.

🎯 Exam Tip: Pay attention to the form of the expression inside the inverse function. Terms like \( 2x^2-1 \) suggest \( \cos 2\theta \) with \( x = \cos\theta \), while \( \frac{1-x^2}{1+x^2} \) suggests \( \cos 2\theta \) with \( x = \tan\theta \).

Question 5. Find the derivative of the following functions with respect to x:
(i) \( \sin^{-1}\left(\frac{1+x^2}{1-x^2}\right) + \cos^{-1}\left(\frac{1+x^2}{1-x^2}\right) \) Hint: \( \sin^{-1}\theta+\cos^{-1}\theta = \frac{\pi}{2} \)
(ii) \( \cos^{-1} (2x) + 2 \cos^{-1} (\sqrt{1-4x^2}) \) Hint: Put \( 2x = \cos\theta \)
Answer:
(i) Let \( y = \sin^{-1}\left(\frac{1+x^2}{1-x^2}\right) + \cos^{-1}\left(\frac{1+x^2}{1-x^2}\right) \).
We know a standard inverse trigonometric identity: \( \sin^{-1}A + \cos^{-1}A = \frac{\pi}{2} \) for \( -1 \le A \le 1 \).
Here, \( A = \frac{1+x^2}{1-x^2} \).
So, \( y = \frac{\pi}{2} \).
To find the derivative, we differentiate \( y \) with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) \)
Since \( \frac{\pi}{2} \) is a constant, its derivative is 0.
\( \frac{dy}{dx} = 0 \)
(ii) Let \( y = \cos^{-1} (2x) + 2 \cos^{-1} (\sqrt{1-4x^2}) \).
From the hint, we substitute \( 2x = \cos\theta \). This means \( \theta = \cos^{-1}(2x) \).
So, \( y = \cos^{-1}(\cos\theta) + 2 \cos^{-1} (\sqrt{1-\cos^2\theta}) \)
\( y = \theta + 2 \cos^{-1} (\sqrt{\sin^2\theta}) \)
\( y = \theta + 2 \cos^{-1} (\sin\theta) \)
We know that \( \sin\theta = \cos\left(\frac{\pi}{2}-\theta\right) \).
\( y = \theta + 2 \cos^{-1} \left(\cos\left(\frac{\pi}{2}-\theta\right)\right) \)
\( y = \theta + 2\left(\frac{\pi}{2}-\theta\right) \)
\( y = \theta + \pi - 2\theta \)
\( y = \pi - \theta \)
Now, we substitute back \( \theta = \cos^{-1}(2x) \).
\( y = \pi - \cos^{-1}(2x) \)
Next, we differentiate \( y \) with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}(\pi - \cos^{-1}(2x)) \)
\( \frac{dy}{dx} = 0 - \left(-\frac{1}{\sqrt{1-(2x)^2}}\right) \cdot \frac{d}{dx}(2x) \)
\( \frac{dy}{dx} = \frac{1}{\sqrt{1-4x^2}} \cdot 2 \)
\( \frac{dy}{dx} = \frac{2}{\sqrt{1-4x^2}} \)
In simple words: The first part simplifies to a constant using a basic inverse trigonometric identity, so its derivative is zero. For the second part, we use a substitution to transform the expression, making it much easier to differentiate after simplification.

🎯 Exam Tip: Always look for opportunities to apply fundamental inverse trigonometric identities or substitutions. They are powerful tools for simplifying complex expressions before differentiation, saving time and reducing error.

Question 6. Find the derivative of the following functions with respect to x:
(i) \( \tan^{-1}\left(\frac{a+x}{1-ax}\right) \) Hint: Put \( x = \tan\theta, a = \tan\alpha \)
(ii) \( \tan^{-1}\left(\frac{2^x+1}{1-4^x}\right) \) Hint: Put \( 2^x = \tan\theta \)
Answer:
(i) Let \( y = \tan^{-1}\left(\frac{a+x}{1-ax}\right) \).
From the hint, we substitute \( x = \tan\theta \) and \( a = \tan\alpha \). This means \( \theta = \tan^{-1}x \) and \( \alpha = \tan^{-1}a \).
So, \( y = \tan^{-1}\left(\frac{\tan\alpha+\tan\theta}{1-\tan\alpha\tan\theta}\right) \)
We know the trigonometric identity \( \tan(\alpha+\theta) = \frac{\tan\alpha+\tan\theta}{1-\tan\alpha\tan\theta} \).
\( y = \tan^{-1}(\tan(\alpha+\theta)) \)
\( y = \alpha+\theta \)
Now, we substitute back \( \alpha = \tan^{-1}a \) and \( \theta = \tan^{-1}x \).
\( y = \tan^{-1}a + \tan^{-1}x \)
Next, we differentiate \( y \) with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}(\tan^{-1}a + \tan^{-1}x) \)
Since \( \tan^{-1}a \) is a constant, its derivative is 0.
\( \frac{dy}{dx} = 0 + \frac{1}{1+x^2} \)
\( \frac{dy}{dx} = \frac{1}{1+x^2} \)
(ii) Let \( y = \tan^{-1}\left(\frac{2^x+1}{1-4^x}\right) \).
We interpret the question as \( \tan^{-1}\left(\frac{2 \cdot 2^x}{1-(2^x)^2}\right) \) based on the provided solution steps. The term `2^x+1` in the OCR seems to be a corrupted version of `2 \cdot 2^x`.
From the hint, we substitute \( 2^x = \tan\theta \). This means \( \theta = \tan^{-1}(2^x) \).
So, \( y = \tan^{-1}\left(\frac{2\tan\theta}{1-\tan^2\theta}\right) \) (where \( 4^x = (2^x)^2 = \tan^2\theta \))
We know the trigonometric identity \( \tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta} \).
\( y = \tan^{-1}(\tan 2\theta) \)
\( y = 2\theta \)
Now, we substitute back \( \theta = \tan^{-1}(2^x) \).
\( y = 2\tan^{-1}(2^x) \)
Next, we differentiate \( y \) with respect to \( x \). We use the chain rule.
\( \frac{dy}{dx} = \frac{d}{dx}(2\tan^{-1}(2^x)) \)
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+(2^x)^2} \cdot \frac{d}{dx}(2^x) \)
We know that \( \frac{d}{dx}(a^x) = a^x \log_e a \). So, \( \frac{d}{dx}(2^x) = 2^x \log_e 2 \).
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1+4^x} \cdot (2^x \log_e 2) \)
\( \frac{dy}{dx} = \frac{2^{x+1} \log_e 2}{1+4^x} \)
In simple words: Both parts use substitution and a trigonometric identity to simplify the inverse tangent function. The first part uses the sum formula for tangent. The second part uses the double angle formula for tangent, treating \( 2^x \) as the tangent of an angle. After simplification, we differentiate using standard rules, including the chain rule for the second part.

🎯 Exam Tip: For expressions like \( \tan^{-1}\left(\frac{A+B}{1-AB}\right) \), think of the sum formula \( \tan(A+B) \). For forms like \( \frac{2A}{1-A^2} \), consider the double angle formula \( \tan 2A \).

Question 7. Find the derivative of the following functions with respect to x:
(i) \( \sin^{-1}\left\{2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right\} \)
(ii) \( \cot^{-1}(\sqrt{1+x^2} + x) \)
Answer:
(i) Let \( y = \sin\left\{2\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right\} \).
We substitute \( x = \cos\theta \). This means \( \theta = \cos^{-1}x \).
So, \( y = \sin\left\{2\tan^{-1}\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\right\} \)
We know the half-angle trigonometric identities: \( 1-\cos\theta = 2\sin^2\left(\frac{\theta}{2}\right) \) and \( 1+\cos\theta = 2\cos^2\left(\frac{\theta}{2}\right) \).
\( y = \sin\left\{2\tan^{-1}\sqrt{\frac{2\sin^2(\theta/2)}{2\cos^2(\theta/2)}}\right\} \)
\( y = \sin\left\{2\tan^{-1}\sqrt{\tan^2\left(\frac{\theta}{2}\right)}\right\} \)
\( y = \sin\left\{2\tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right)\right\} \)
\( y = \sin\left\{2 \cdot \frac{\theta}{2}\right\} \)
\( y = \sin\theta \)
Now, we substitute back \( \theta = \cos^{-1}x \). Since \( x = \cos\theta \), we know \( \sin\theta = \sqrt{1-\cos^2\theta} = \sqrt{1-x^2} \).
\( y = \sqrt{1-x^2} \)
Next, we differentiate \( y \) with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}(\sqrt{1-x^2}) \)
\( \frac{dy}{dx} = \frac{1}{2\sqrt{1-x^2}} \cdot \frac{d}{dx}(1-x^2) \)
\( \frac{dy}{dx} = \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) \)
\( \frac{dy}{dx} = \frac{-x}{\sqrt{1-x^2}} \)
(ii) Let \( y = \cot^{-1}(\sqrt{1+x^2} + x) \).
We substitute \( x = \tan\theta \). This means \( \theta = \tan^{-1}x \).
So, \( y = \cot^{-1}(\sqrt{1+\tan^2\theta} + \tan\theta) \)
We know the identity \( 1+\tan^2\theta = \sec^2\theta \).
\( y = \cot^{-1}(\sqrt{\sec^2\theta} + \tan\theta) \)
\( y = \cot^{-1}(\sec\theta + \tan\theta) \)
Now, we express \( \sec\theta \) and \( \tan\theta \) in terms of \( \sin\theta \) and \( \cos\theta \).
\( y = \cot^{-1}\left(\frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\right) \)
\( y = \cot^{-1}\left(\frac{1+\sin\theta}{\cos\theta}\right) \)
To simplify this, we use the identities \( 1+\sin\theta = 1+\cos\left(\frac{\pi}{2}-\theta\right) = 2\cos^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right) \) and \( \cos\theta = \sin\left(\frac{\pi}{2}-\theta\right) = 2\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right) \).
\( y = \cot^{-1}\left(\frac{2\cos^2\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{2\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}\right) \)
\( y = \cot^{-1}\left(\frac{\cos\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}{\sin\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}\right) \)
\( y = \cot^{-1}\left(\cot\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right) \)
\( y = \frac{\pi}{4}-\frac{\theta}{2} \)
Now, we substitute back \( \theta = \tan^{-1}x \).
\( y = \frac{\pi}{4}-\frac{1}{2}\tan^{-1}x \)
Next, we differentiate \( y \) with respect to \( x \).
\( \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4}-\frac{1}{2}\tan^{-1}x\right) \)
\( \frac{dy}{dx} = 0 - \frac{1}{2} \cdot \frac{1}{1+x^2} \)
\( \frac{dy}{dx} = -\frac{1}{2(1+x^2)} \)
In simple words: For the first part, we use a cosine substitution and half-angle identities to simplify the inner inverse tangent, reducing the whole expression to a simple square root function which is then differentiated. For the second part, a tangent substitution is used, followed by expressing secant and tangent in terms of sine and cosine. This simplifies the expression to a cotangent function, which allows for easy differentiation after converting back to inverse tangent.

🎯 Exam Tip: When dealing with \( \sqrt{\frac{1-x}{1+x}} \), the substitution \( x=\cos\theta \) is almost always the most effective. For \( \sqrt{1+x^2} \), consider \( x=\tan\theta \) or \( x=\cot\theta \).