CBSE Class 12 Mathematics HOTs Continuity And Differentiability Set 07

Question. If \( y = \sin^{-1}(x\sqrt{1-x} - \sqrt{x}\sqrt{1-x^2}) \) and \( 0 < x < 1 \), then find \( \frac{dy}{dx} \).
Answer: We have, \( y = \sin^{-1}(x\sqrt{1-x} - \sqrt{x}\sqrt{1-x^2}) \)
\( \Rightarrow y = \sin^{-1}(x\sqrt{1-(\sqrt{x})^2} - \sqrt{x}\sqrt{1-x^2}) \)
\( \Rightarrow y = \sin^{-1} x - \sin^{-1} \sqrt{x} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x}) \)
\( = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{2\sqrt{x - x^2}} \)

Question. Show that the function \( f(x) = |x - 3| \), \( x \in \mathbb{R} \), is continuous but not differentiable at \( x = 3 \).
Answer: \( f(x) = |x - 3| = \begin{cases} x - 3, & \text{if } x \ge 3 \\ -(x - 3), & \text{if } x < 3 \end{cases} \)
We have, \( f(3) = |3 - 3| = 0 \)
\( \lim_{x \to 3^+} f(x) = \lim_{h \to 0} f(3+h) = \lim_{h \to 0} (3+h-3) = \lim_{h \to 0} h = 0 \)
\( \lim_{x \to 3^-} f(x) = \lim_{h \to 0} f(3-h) = \lim_{h \to 0} [-(3-h-3)] = \lim_{h \to 0} h = 0 \)
\( \because \lim_{x \to 3^+} f(x) = \lim_{x \to 3^-} f(x) = f(3) = 0 \),
So, \( f(x) \) is continuous at \( x = 3 \).
Now, \( Rf'(3) = \lim_{h \to 0} \frac{f(3+h) - f(3)}{h} = \lim_{h \to 0} \frac{(3+h-3) - 0}{h} = \lim_{h \to 0} \frac{h}{h} = 1 \)
And \( Lf'(3) = \lim_{h \to 0} \frac{f(3-h) - f(3)}{-h} = \lim_{h \to 0} \frac{[-(3-h-3)] - 0}{-h} = \lim_{h \to 0} \frac{h}{-h} = -1 \)
Thus, \( Rf'(3) \ne Lf'(3) \)
\( \therefore f(x) \) is not differentiable at \( x = 3 \).

Question. If \( \sin y = x \sin(a + y) \), then prove that \( \frac{dy}{dx} = \frac{\sin^2(a + y)}{\sin a} \).
Answer: We have, \( \sin y = x \sin(a + y) \)
\( \Rightarrow x = \frac{\sin y}{\sin(a + y)} \)
Differentiating w.r.t. \( y \), we get
\( \frac{dx}{dy} = \frac{\sin(a + y)\frac{d}{dy}(\sin y) - \sin y\frac{d}{dy}(\sin(a + y))}{\sin^2(a + y)} \)
\( \Rightarrow \frac{dx}{dy} = \frac{\sin(a + y)\cos y - \sin y\cos(a + y)}{\sin^2(a + y)} \)
\( \Rightarrow \frac{dx}{dy} = \frac{\sin[(a + y) - y]}{\sin^2(a + y)} = \frac{\sin a}{\sin^2(a + y)} \)
\( \therefore \frac{dy}{dx} = \frac{\sin^2(a + y)}{\sin a} \)

Question. Differentiate \( \tan^{-1} \left[ \frac{\sqrt{1+x^2} - 1}{x} \right] \) with respect to \( x \).
Answer: Let \( y = \tan^{-1} \left[ \frac{\sqrt{1+x^2} - 1}{x} \right] \)
Put \( x = \tan\theta \Rightarrow \theta = \tan^{-1} x \)
\( \therefore y = \tan^{-1} \left[ \frac{\sqrt{1+\tan^2\theta} - 1}{\tan\theta} \right] = \tan^{-1} \left[ \frac{\sec\theta - 1}{\tan\theta} \right] \)
\( \Rightarrow y = \tan^{-1} \left[ \frac{\frac{1}{\cos\theta} - 1}{\frac{\sin\theta}{\cos\theta}} \right] = \tan^{-1} \left[ \frac{1 - \cos\theta}{\sin\theta} \right] \)
\( \Rightarrow y = \tan^{-1} \left[ \frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \right] = \tan^{-1} \left[ \tan\frac{\theta}{2} \right] = \frac{\theta}{2} = \frac{1}{2}\tan^{-1}x \)
\( \therefore \frac{dy}{dx} = \frac{1}{2(1+x^2)} \)

Question. If \( x\sqrt{1+y} + y\sqrt{1+x} = 0 \) for \( x \ne y \), prove that \( \frac{dy}{dx} = \frac{-1}{(1+x)^2} \).
Answer: We have, \( x\sqrt{1+y} + y\sqrt{1+x} = 0 \)
\( \Rightarrow x\sqrt{1+y} = -y\sqrt{1+x} \)
Squaring both sides, we get
\( x^2(1+y) = y^2(1+x) \)
\( \Rightarrow x^2 + x^2y = y^2 + xy^2 \)
\( \Rightarrow x^2 - y^2 + x^2y - xy^2 = 0 \)
\( \Rightarrow (x-y)(x+y) + xy(x-y) = 0 \)
\( \Rightarrow (x-y)(x+y+xy) = 0 \)
Since \( x \ne y \), we have:
\( x + y + xy = 0 \Rightarrow y(1+x) = -x \Rightarrow y = \frac{-x}{1+x} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = \frac{(1+x)\frac{d}{dx}(-x) - (-x)\frac{d}{dx}(1+x)}{(1+x)^2} \)
\( = \frac{(1+x)(-1) + x(1)}{(1+x)^2} = \frac{-1-x+x}{(1+x)^2} = \frac{-1}{(1+x)^2} \)

Question. If \( y = a\sin x + b\cos x \), prove that \( y^2 + \left( \frac{dy}{dx} \right)^2 = a^2 + b^2 \).
Answer: Here, \( y = a\sin x + b\cos x \)
\( \Rightarrow \frac{dy}{dx} = a\cos x - b\sin x \)
Now, \( \text{L.H.S.} = y^2 + \left( \frac{dy}{dx} \right)^2 \)
\( = (a\sin x + b\cos x)^2 + (a\cos x - b\sin x)^2 \)
\( = a^2\sin^2 x + b^2\cos^2 x + 2ab\sin x\cos x + a^2\cos^2 x + b^2\sin^2 x - 2ab\sin x\cos x \)
\( = a^2(\sin^2 x + \cos^2 x) + b^2(\cos^2 x + \sin^2 x) \)
\( = a^2(1) + b^2(1) = a^2 + b^2 = \text{R.H.S.} \)

Question. Show that the function defined as follows, is continuous at \( x = 2 \), but not differentiable.
\( f(x) = \begin{cases} 3x - 2, & 0 < x < 1 \\ 2x^2 - x, & 1 \le x \le 2 \\ 5x - 4, & x > 2 \end{cases} \)
Answer: L.H.L. \( = \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x^2 - x) = 2(2)^2 - 2 = 6 \)
R.H.L. \( = \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (5x - 4) = 5(2) - 4 = 6 \)
Also, \( f(2) = 2(2)^2 - 2 = 6 \)
As, \( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \),
\( \therefore f(x) \) is continuous at \( x = 2 \).
Test of differentiability:
We have, \( Lf'(2) = \lim_{h \to 0} \frac{f(2-h) - f(2)}{-h} \)
\( = \lim_{h \to 0} \frac{2(2-h)^2 - (2-h) - 6}{-h} \)
\( = \lim_{h \to 0} \frac{2(4 - 4h + h^2) - 2 + h - 6}{-h} \)
\( = \lim_{h \to 0} \frac{8 - 8h + 2h^2 - 8 + h}{-h} \)
\( = \lim_{h \to 0} \frac{2h^2 - 7h}{-h} = \lim_{h \to 0} \frac{-h(-2h + 7)}{-h} = \lim_{h \to 0} (-2h + 7) = 7 \)
\( Rf'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h} \)
\( = \lim_{h \to 0} \frac{5(2+h) - 4 - 6}{h} \)
\( = \lim_{h \to 0} \frac{10 + 5h - 10}{h} = \lim_{h \to 0} \frac{5h}{h} = 5 \)
\( \because Lf'(2) \ne Rf'(2) \),
Hence, \( f(x) \) is not differentiable at \( x = 2 \).

Question. If \( y = \cos^{-1}\left[ \frac{3x + 4\sqrt{1-x^2}}{5} \right] \), find \( \frac{dy}{dx} \).
Answer: We have, \( y = \cos^{-1}\left[ \frac{3x + 4\sqrt{1-x^2}}{5} \right] \)
Putting \( x = \sin\theta \Rightarrow \theta = \sin^{-1}x \), we get:
\( y = \cos^{-1}\left[ \frac{3\sin\theta + 4\cos\theta}{5} \right] \)
\( \Rightarrow y = \cos^{-1}\left[ \frac{3}{5}\sin\theta + \frac{4}{5}\cos\theta \right] \)
Let \( \frac{3}{5} = \sin\alpha \) and \( \frac{4}{5} = \cos\alpha \)
\( \Rightarrow y = \cos^{-1}[\sin\alpha\sin\theta + \cos\alpha\cos\theta] \)
\( \Rightarrow y = \cos^{-1}[\cos(\alpha - \theta)] \Rightarrow y = \alpha - \theta \)
\( \Rightarrow y = \alpha - \sin^{-1}x \)
Differentiating w.r.t. \( x \), we get:
\( \frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}} \)

Question. If \( y = \cos^{-1}\left[ \frac{2x - 3\sqrt{1-x^2}}{\sqrt{13}} \right] \), find \( \frac{dy}{dx} \).
Answer: We have, \( y = \cos^{-1}\left[ \frac{2x - 3\sqrt{1-x^2}}{\sqrt{13}} \right] \)
Putting \( x = \sin\theta \Rightarrow \theta = \sin^{-1}x \), we get:
\( y = \cos^{-1}\left[ \frac{2\sin\theta - 3\cos\theta}{\sqrt{13}} \right] \)
\( \Rightarrow y = \cos^{-1}\left[ \frac{2}{\sqrt{13}}\sin\theta - \frac{3}{\sqrt{13}}\cos\theta \right] \)
Let \( \frac{2}{\sqrt{13}} = \sin\alpha \) and \( \frac{3}{\sqrt{13}} = \cos\alpha \)
\( \Rightarrow y = \cos^{-1}[\sin\alpha\sin\theta - \cos\alpha\cos\theta] \)
\( \Rightarrow y = \cos^{-1}[-(\cos\alpha\cos\theta - \sin\alpha\sin\theta)] \)
\( \Rightarrow y = \cos^{-1}[-\cos(\alpha + \theta)] \)
\( \Rightarrow y = \pi - (\alpha + \theta) = \pi - \alpha - \sin^{-1}x \)
Differentiating w.r.t. \( x \), we get:
\( \frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}} \)

Question. Find \( \frac{dy}{dx} \) if \( (x^2 + y^2)^2 = xy \).
Answer: We have, \( (x^2 + y^2)^2 = xy \)
Differentiating w.r.t. \( x \), we get:
\( 2(x^2 + y^2)\left[ 2x + 2y\frac{dy}{dx} \right] = x\frac{dy}{dx} + y \)
\( \Rightarrow 4x(x^2 + y^2) + 4y(x^2 + y^2)\frac{dy}{dx} = x\frac{dy}{dx} + y \)
\( \Rightarrow \left[ 4y(x^2 + y^2) - x \right]\frac{dy}{dx} = y - 4x(x^2 + y^2) \)
\( \Rightarrow \frac{dy}{dx} = \frac{y - 4x^3 - 4xy^2}{4x^2y + 4y^3 - x} \)

Question. If \( y = \cot^{-1}\left[ \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}} \right] \), find \( \frac{dy}{dx} \).
Answer: We have, \( y = \cot^{-1}\left[ \frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}} \right] \)
We know that \( 1 + \sin x = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2 \) and \( 1 - \sin x = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2 \)
\( \therefore \sqrt{1+\sin x} = \cos\frac{x}{2} + \sin\frac{x}{2} \) and \( \sqrt{1-\sin x} = \cos\frac{x}{2} - \sin\frac{x}{2} \)
Substituting these values, we get:
\( y = \cot^{-1}\left[ \frac{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right) + \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right) - \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)} \right] \)
\( \Rightarrow y = \cot^{-1}\left[ \frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} \right] \)
\( \Rightarrow y = \cot^{-1}\left[ \cot\frac{x}{2} \right] \)
\( \Rightarrow y = \frac{x}{2} \)
Differentiating w.r.t. \( x \), we get:
\( \frac{dy}{dx} = \frac{1}{2} \)

Question. Differentiate \( \tan^{-1}\left[ \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right] \) w.r.t. \( x \).
Answer: Let \( y = \tan^{-1}\left[ \frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} \right] \)
Putting \( x = \cos 2\theta \Rightarrow \theta = \frac{1}{2}\cos^{-1}x \), we get:
\( y = \tan^{-1}\left[ \frac{\sqrt{1+\cos 2\theta} - \sqrt{1-\cos 2\theta}}{\sqrt{1+\cos 2\theta} + \sqrt{1-\cos 2\theta}} \right] \)
\( \Rightarrow y = \tan^{-1}\left[ \frac{\sqrt{2\cos^2\theta} - \sqrt{2\sin^2\theta}}{\sqrt{2\cos^2\theta} + \sqrt{2\sin^2\theta}} \right] \)
\( \Rightarrow y = \tan^{-1}\left[ \frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta} \right] \)
\( \Rightarrow y = \tan^{-1}\left[ \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} \right] \)
Dividing numerator and denominator by \( \cos\theta \), we get:
\( y = \tan^{-1}\left[ \frac{1 - \tan\theta}{1 + \tan\theta} \right] = \tan^{-1}\left[ \tan\left(\frac{\pi}{4} - \theta\right) \right] \)
\( \Rightarrow y = \frac{\pi}{4} - \theta \)
\( \Rightarrow y = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x \)
Differentiating w.r.t. \( x \), we get:
\( \frac{dy}{dx} = 0 - \frac{1}{2}\left( \frac{-1}{\sqrt{1-x^2}} \right) = \frac{1}{2\sqrt{1-x^2}} \)

Question. If \( xy + y^2 = \tan x + y \), find \( \frac{dy}{dx} \).
Answer: Given, \( xy + y^2 = \tan x + y \)
Differentiating w.r.t. \( x \) on both sides, we get:
\( x\frac{dy}{dx} + y + 2y\frac{dy}{dx} = \sec^2 x + \frac{dy}{dx} \)
\( \Rightarrow x\frac{dy}{dx} + 2y\frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x - y \)
\( \Rightarrow (x + 2y - 1)\frac{dy}{dx} = \sec^2 x - y \)
\( \Rightarrow \frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1} \)

Question. Differentiate \( \sin^{-1}\left[ \frac{5x + 12\sqrt{1-x^2}}{13} \right] \) w.r.t. \( x \).
Answer: Let \( y = \sin^{-1}\left[ \frac{5x + 12\sqrt{1-x^2}}{13} \right] \)
Putting \( x = \sin\theta \Rightarrow \theta = \sin^{-1}x \), we get:
\( y = \sin^{-1}\left[ \frac{5\sin\theta + 12\cos\theta}{13} \right] \)
\( \Rightarrow y = \sin^{-1}\left[ \frac{5}{13}\sin\theta + \frac{12}{13}\cos\theta \right] \)
Let \( \frac{5}{13} = \cos\alpha \) and \( \frac{12}{13} = \sin\alpha \), we get:
\( y = \sin^{-1}[\sin\theta\cos\alpha + \cos\theta\sin\alpha] \)
\( \Rightarrow y = \sin^{-1}[\sin(\theta + \alpha)] \)
\( \Rightarrow y = \theta + \alpha = \sin^{-1}x + \alpha \)
Differentiating w.r.t. \( x \), we get:
\( \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} \)

Question. If \( y = \frac{x\cos^{-1}x}{\sqrt{1-x^2}} - \log\sqrt{1-x^2} \), then prove that \( \frac{dy}{dx} = \frac{\cos^{-1}x}{(1-x^2)^{3/2}} \).
Answer: Here, \( y = \frac{x\cos^{-1}x}{\sqrt{1-x^2}} - \log(1-x^2)^{1/2} = \frac{x\cos^{-1}x}{\sqrt{1-x^2}} - \frac{1}{2}\log(1-x^2) \)
Differentiating w.r.t. \( x \), we get:
\( \frac{dy}{dx} = \frac{\sqrt{1-x^2}\frac{d}{dx}(x\cos^{-1}x) - x\cos^{-1}x \frac{d}{dx}\left(\sqrt{1-x^2}\right)}{1-x^2} - \frac{1}{2}\frac{1}{1-x^2}(-2x) \)
\( = \frac{\sqrt{1-x^2}\left[\cos^{-1}x - \frac{x}{\sqrt{1-x^2}}\right] - x\cos^{-1}x \left(\frac{-x}{\sqrt{1-x^2}}\right)}{1-x^2} + \frac{x}{1-x^2} \)
\( = \frac{\sqrt{1-x^2}\cos^{-1}x - x + \frac{x^2\cos^{-1}x}{\sqrt{1-x^2}}}{1-x^2} + \frac{x}{1-x^2} \)
\( = \frac{\sqrt{1-x^2}\cos^{-1}x + \frac{x^2\cos^{-1}x}{\sqrt{1-x^2}}}{1-x^2} \)
\( = \frac{\left(1-x^2\right)\cos^{-1}x + x^2\cos^{-1}x}{(1-x^2)\sqrt{1-x^2}} \)
\( = \frac{\cos^{-1}x}{(1-x^2)^{3/2}} \)

Question. If \( e^x + e^y = e^{x+y} \), prove that \( \frac{dy}{dx} + e^{y-x} = 0 \).
Answer: Given \( e^x + e^y = e^{x+y} \Rightarrow 1 + e^{y-x} = e^y \) ...(1)
Differentiating (1) w.r.t. \( x \), we get
\( e^{y-x} \cdot \frac{d}{dx}(y-x) = e^y \frac{dy}{dx} \)
\( \Rightarrow e^{y-x} \left(\frac{dy}{dx} - 1\right) = e^y \frac{dy}{dx} \)
\( \Rightarrow \frac{dy}{dx} (e^{y-x} - e^y) = e^{y-x} \)
\( \Rightarrow \frac{dy}{dx} (-1) = e^{y-x} \) [Using (1)]
\( \Rightarrow \frac{dy}{dx} + e^{y-x} = 0 \)

Question. If \( y = \tan^{-1} \left(\frac{a}{x}\right) + \log \sqrt{\frac{x-a}{x+a}} \), prove that \( \frac{dy}{dx} = \frac{2a^3}{x^4 - a^4} \).
Answer: Here, \( y = \tan^{-1}\left(\frac{a}{x}\right) + \log \sqrt{\frac{x-a}{x+a}} \)
\( = \tan^{-1}\left(\frac{a}{x}\right) + \frac{1}{2}\log \left(\frac{x-a}{x+a}\right) \)
\( = \tan^{-1}\left(\frac{a}{x}\right) + \frac{1}{2} [\log(x-a) - \log(x+a)] \)
Differentiating w.r.t \( x \), we get
\( \frac{dy}{dx} = \frac{1}{1 + \frac{a^2}{x^2}} \frac{d}{dx}\left(\frac{a}{x}\right) + \frac{1}{2}\left[\frac{1}{x-a} - \frac{1}{x+a}\right] \)
\( = \frac{x^2}{x^2 + a^2} \cdot a \left(-\frac{1}{x^2}\right) + \frac{1}{2}\left[\frac{(x+a) - (x-a)}{x^2 - a^2}\right] \)
\( = \frac{-a}{x^2 + a^2} + \frac{a}{x^2 - a^2} \)
\( = \frac{-a(x^2 - a^2) + a(x^2 + a^2)}{x^4 - a^4} \)
\( = \frac{2a^3}{x^4 - a^4} \)

Question. If \( \log(\sqrt{1+x^2} - x) = y\sqrt{1+x^2} \), show that \( (1+x^2)\frac{dy}{dx} + xy + 1 = 0 \).
Answer: We have, \( \log(\sqrt{1+x^2} - x) = y\sqrt{1+x^2} \)
Differentiating w.r.t. \( x \), we get
\( \frac{1}{\sqrt{1+x^2}-x} \left[ \frac{1}{2\sqrt{1+x^2}} \cdot 2x - 1 \right] = \frac{dy}{dx} \sqrt{1+x^2} + y \cdot \frac{x}{\sqrt{1+x^2}} \)
\( \Rightarrow \frac{1}{\sqrt{1+x^2}-x} \cdot \frac{x-\sqrt{1+x^2}}{\sqrt{1+x^2}} = (1+x^2)\frac{dy}{dx} + xy \cdot \frac{1}{\sqrt{1+x^2}} \)
\( \Rightarrow \frac{-1}{\sqrt{1+x^2}} = \frac{(1+x^2)\frac{dy}{dx} + xy}{\sqrt{1+x^2}} \)
\( \Rightarrow (1+x^2)\frac{dy}{dx} + xy + 1 = 0 \)

Question. If \( y = \sqrt{x^2+1} - \log \left[ \frac{1}{x} + \sqrt{1+\frac{1}{x^2}} \right] \), find \( \frac{dy}{dx} \).
Answer: We have, \( y = \sqrt{x^2+1} - \log \left[ \frac{1}{x} + \sqrt{1 + \frac{1}{x^2}} \right] \)
\( \Rightarrow y = \sqrt{x^2+1} - \log \left[ \frac{1 + \sqrt{x^2+1}}{x} \right] \)
\( \Rightarrow y = \sqrt{x^2+1} - \log(1 + \sqrt{x^2+1}) + \log x \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = \frac{2x}{2\sqrt{x^2+1}} - \frac{1}{1+\sqrt{x^2+1}} \left( \frac{2x}{2\sqrt{x^2+1}} \right) + \frac{1}{x} \)
\( \Rightarrow \frac{dy}{dx} = \frac{x}{\sqrt{x^2+1}} - \frac{x}{\sqrt{x^2+1}(1+\sqrt{x^2+1})} + \frac{1}{x} \)
\( = \frac{x(1+\sqrt{x^2+1}) - x}{\sqrt{x^2+1}(1+\sqrt{x^2+1})} + \frac{1}{x} = \frac{x}{1+\sqrt{x^2+1}} + \frac{1}{x} \)
\( = \frac{x^2 + 1 + \sqrt{x^2+1}}{x(1+\sqrt{x^2+1})} = \frac{\sqrt{x^2+1}(\sqrt{x^2+1} + 1)}{x(1+\sqrt{x^2+1})} \)
\( = \frac{\sqrt{x^2+1}}{x} \)

Question. If \( y = \log \sqrt{\frac{1-\cos 2x}{1+\cos 2x}} \), then show that \( \frac{dy}{dx} = 2 \csc 2x \).
Answer: We have, \( y = \log \sqrt{\frac{1-\cos 2x}{1+\cos 2x}} \)
\( \Rightarrow y = \log \sqrt{\frac{2\sin^2 x}{2\cos^2 x}} = \log(\tan x) \)
\( \therefore \frac{dy}{dx} = \frac{1}{\tan x} \times \sec^2 x \)
\( = \frac{\cos x}{\sin x} \times \frac{1}{\cos^2 x} = \frac{2}{2\sin x \cos x} = \frac{2}{\sin 2x} = 2 \csc 2x \)

Logarithmic Differentiation

Question. Differentiate \( x^{\sin x} + (\sin x)^{\cos x} \) with respect to \( x \).
Answer: Let \( y = x^{\sin x} + (\sin x)^{\cos x} \)
\( \Rightarrow y = e^{\sin x \log x} + e^{\cos x \log \sin x} \)
Differentiating both sides w.r.t. \( x \), we get
\( \frac{dy}{dx} = e^{\sin x \log x} \left[ \sin x \cdot \frac{1}{x} + \log x \cdot \cos x \right] + e^{\cos x \log \sin x} \left[ \cos x \cdot \frac{\cos x}{\sin x} + \log \sin x \cdot (-\sin x) \right] \)
\( \Rightarrow \frac{dy}{dx} = x^{\sin x} \left[ \frac{\sin x}{x} + \log x \cdot \cos x \right] + (\sin x)^{\cos x} \left[ \frac{\cos^2 x}{\sin x} - \sin x \log \sin x \right] \)

Question. If \( y = (\sin x)^x + \sin^{-1} \sqrt{x} \), then find \( \frac{dy}{dx} \).
Answer: Here, \( y = (\sin x)^x + \sin^{-1} \sqrt{x} \)
\( \Rightarrow y = e^{x \log \sin x} + \sin^{-1} \sqrt{x} \)
\( \therefore \frac{dy}{dx} = e^{x \log \sin x} \left[ x \cdot \frac{\cos x}{\sin x} + \log \sin x \right] + \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}} \)
\( \Rightarrow \frac{dy}{dx} = (\sin x)^x (\log \sin x + x \cot x) + \frac{1}{2\sqrt{x - x^2}} \)

Question. If \( x^m y^n = (x+y)^{m+n} \), prove that \( \frac{dy}{dx} = \frac{y}{x} \).
Answer: Given \( x^m y^n = (x+y)^{m+n} \)
Taking log on both sides, we get
\( \log x^m + \log y^n = (m+n) \log(x+y) \)
\( \Rightarrow m \log x + n \log y = (m+n) \log(x+y) \)
Differentiating w.r.t. \( x \), we get
\( \frac{m}{x} + \frac{n}{y} \frac{dy}{dx} = \frac{m+n}{x+y} \left( 1 + \frac{dy}{dx} \right) \)
\( \Rightarrow \left( \frac{n}{y} - \frac{m+n}{x+y} \right) \frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x} \)
\( \Rightarrow \left[ \frac{n(x+y) - y(m+n)}{y(x+y)} \right] \frac{dy}{dx} = \frac{x(m+n) - m(x+y)}{x(x+y)} \)
\( \Rightarrow \left[ \frac{nx + ny - my - ny}{y(x+y)} \right] \frac{dy}{dx} = \frac{mx + nx - mx - my}{x(x+y)} \)
\( \Rightarrow \left[ \frac{nx - my}{y(x+y)} \right] \frac{dy}{dx} = \frac{nx - my}{x(x+y)} \)
\( \therefore \frac{dy}{dx} = \frac{y}{x} \)

Question. If \( (x-y) \cdot e^{\frac{x}{x-y}} = a \), prove that \( y\frac{dy}{dx} + x = 2y \).
Answer: Here, \( (x-y) \cdot e^{\frac{x}{x-y}} = a \)
Taking log on both sides, we get
\( \log \left\{ (x-y) \cdot e^{\frac{x}{x-y}} \right\} = \log a \)
\( \Rightarrow \log(x-y) + \frac{x}{x-y} = \log a \)
Differentiating w.r.t. \( x \), we get
\( \frac{1}{x-y} \left( 1 - \frac{dy}{dx} \right) + \frac{(x-y) \cdot 1 - x \left( 1 - \frac{dy}{dx} \right)}{(x-y)^2} = 0 \)
\( \Rightarrow (x-y) \left( 1 - \frac{dy}{dx} \right) + x - y - x + x \frac{dy}{dx} = 0 \)
\( \Rightarrow x - 2y + y \frac{dy}{dx} = 0 \Rightarrow y \frac{dy}{dx} + x = 2y \)

Question. If \( (\tan^{-1} x)^y + y^{\cot x} = 1 \), then find \( \frac{dy}{dx} \).
Answer: Here, \( (\tan^{-1} x)^y + y^{\cot x} = 1 \)
\( \Rightarrow u + v = 1 \) where \( u = (\tan^{-1} x)^y \) and \( v = y^{\cot x} \)
Differentiating w.r.t. \( x \), we get
\( \frac{du}{dx} + \frac{dv}{dx} = 0 \) ...(1)
Now, \( u = (\tan^{-1} x)^y \)
\( \Rightarrow \log u = y \log(\tan^{-1} x) \)
Differentiating w.r.t. \( x \), we get
\( \frac{1}{u} \frac{du}{dx} = \frac{dy}{dx} \log(\tan^{-1} x) + y \cdot \frac{1}{\tan^{-1} x} \cdot \frac{1}{1+x^2} \)
\( \Rightarrow \frac{du}{dx} = (\tan^{-1} x)^y \left[ \frac{dy}{dx} \log(\tan^{-1} x) + \frac{y}{(1+x^2)\tan^{-1} x} \right] \) ...(2)
And \( v = y^{\cot x} \)
\( \Rightarrow \log v = \cot x \cdot \log y \)
Differentiating w.r.t. \( x \), we get
\( \frac{1}{v} \frac{dv}{dx} = \cot x \cdot \frac{1}{y} \frac{dy}{dx} - \csc^2 x \cdot \log y \)
\( \Rightarrow \frac{dv}{dx} = y^{\cot x} \left[ \frac{\cot x}{y} \frac{dy}{dx} - \csc^2 x \cdot \log y \right] \) ...(3)
From (1), (2) and (3), we get
\( (\tan^{-1} x)^y \left[ \frac{dy}{dx} \log(\tan^{-1} x) + \frac{y}{(1+x^2)\tan^{-1} x} \right] + y^{\cot x} \left[ \frac{\cot x}{y} \frac{dy}{dx} - \csc^2 x \cdot \log y \right] = 0 \)
\( \Rightarrow \frac{dy}{dx} \left[ (\tan^{-1} x)^y \log(\tan^{-1} x) + y^{\cot x - 1} \cot x \right] = y^{\cot x} \csc^2 x \log y - (\tan^{-1} x)^{y-1} \frac{y}{1+x^2} \)
\( \therefore \frac{dy}{dx} = \frac{y^{\cot x} \csc^2 x \log y - (\tan^{-1} x)^{y-1} \frac{y}{1+x^2}}{(\tan^{-1} x)^y \log(\tan^{-1} x) + y^{\cot x-1} \cot x} \)

Question. Differentiate the following function with respect to \( x \): \( (\log x)^x + x^{\log x} \).
Answer: Let \( y = (\log x)^x + x^{\log x} \)
\( \therefore y = e^{x \log(\log x)} + e^{(\log x)^2} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = e^{x \log(\log x)} \frac{d}{dx} \{ x \log(\log x) \} + e^{(\log x)^2} \frac{d}{dx} \{ (\log x)^2 \} \)
\( = (\log x)^x \left[ x \cdot \frac{1}{\log x} \cdot \frac{1}{x} + \log(\log x) \right] + x^{\log x} \left[ 2 \log x \cdot \frac{1}{x} \right] \)
\( = (\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right] + 2 \left( \frac{\log x}{x} \right) x^{\log x} \)

Question. If \( y^x = e^{y-x} \), prove that \( \frac{dy}{dx} = \frac{(1+\log y)^2}{\log y} \).
Answer: Here \( y^x = e^{y-x} \)
Taking log on both sides, we get
\( x \log y = (y - x)\log e = y - x \)
\( \Rightarrow x(1+\log y) = y \)
\( \Rightarrow x = \frac{y}{1 + \log y} \)
Differentiating w.r.t. \( y \), we get
\( \frac{dx}{dy} = \frac{(1 + \log y) \cdot 1 - y \cdot \frac{1}{y}}{(1 + \log y)^2} = \frac{\log y}{(1 + \log y)^2} \)
\( \Rightarrow \frac{dy}{dx} = \frac{(1 + \log y)^2}{\log y} \)

Question. Differentiate the following with respect to \( x \): \( \sin^{-1} \left[ \frac{2^{x+1} \cdot 3^x}{1 + (36)^x} \right] \).
Answer: Let \( y = \sin^{-1} \left[ \frac{2^{x+1} \cdot 3^x}{1 + (36)^x} \right] \)
\( \Rightarrow y = \sin^{-1} \left[ \frac{2 \cdot 2^x \cdot 3^x}{1 + (36)^x} \right] = \sin^{-1} \left[ \frac{2 \cdot 6^x}{1 + (6^x)^2} \right] \)
Put \( 6^x = \tan \theta \Rightarrow \theta = \tan^{-1} (6^x) \)
\( \dots y = \sin^{-1} \left[ \frac{2 \tan \theta}{1 + \tan^2 \theta} \right] = \sin^{-1}(\sin 2\theta) = 2\theta \)
\( \Rightarrow y = 2\theta = 2 \tan^{-1}(6^x) \)
Now differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1 + (6^x)^2} \frac{d}{dx}(6^x) \)
\( = \frac{2}{1 + (36)^x} \cdot 6^x \log 6 = \frac{2\log 6 \cdot 6^x}{1 + (36)^x} \)

Question. If \( x^y = e^{x-y} \), prove that \( \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} \).
Answer: We have, \( x^y = e^{x-y} \)
Taking log on both sides, we get
\( y \log x = (x - y)\log e = x - y \)
\( \Rightarrow y(1 + \log x) = x \Rightarrow y = \frac{x}{1 + \log x} \)
Differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = \frac{(1 + \log x) \cdot 1 - x \cdot \left(\frac{1}{x}\right)}{(1 + \log x)^2} = \frac{1 + \log x - 1}{(1 + \log x)^2} \)
\( \therefore \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} \)

Question. Find \( \frac{dy}{dx} \), if \( y = \sin^{-1} \left( \frac{2^{x+1}}{1 + 4^x} \right) \).
Answer: We have, \( y = \sin^{-1} \left( \frac{2^{x+1}}{1 + 4^x} \right) \)
\( \Rightarrow y = \sin^{-1} \left( \frac{2 \cdot 2^x}{1 + (2^x)^2} \right) \)
Put \( 2^x = \tan \theta \Rightarrow \theta = \tan^{-1}(2^x) \)
\( \therefore y = \sin^{-1} \left( \frac{2\tan\theta}{1+\tan^2\theta} \right) = \sin^{-1}(\sin 2\theta) = 2\theta = 2\tan^{-1}(2^x) \)
Now differentiating w.r.t. \( x \), we get
\( \frac{dy}{dx} = 2 \cdot \frac{1}{1 + (2^x)^2} \frac{d}{dx}(2^x) \)
\( = \frac{2}{1 + 4^x} \cdot 2^x \log 2 = \frac{2^{x+1} \log 2}{1 + 4^x} \)