CBSE Class 12 Mathematics Continuity and Differentiability VBQs Set 02

Fill in the Blanks

Question. If \( y = \tan^{-1} x + \cot^{-1} x, x \in R \), then \( \frac{dy}{dx} \) is equal to _____________. 
Answer: 0
Solution: We have, \( y = \tan^{-1} x + \cot^{-1} x, x \in R \)
\( \Rightarrow y = \frac{\pi}{2} \)
\( \therefore \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) = 0 \)

Question. If \( \cos (xy) = k \), where \( k \) is a constant and \( xy \neq n\pi, n \in Z \), then \( \frac{dy}{dx} \) is equal to _____________. 
Answer: \( \frac{-y}{x} \)
Solution: We have, \( \cos (xy) = k \)
Diff. w.r.t \( x \), we get
\( -\sin(xy) \times \left\{ x\frac{dy}{dx} + y \right\} = 0 \)
\( \Rightarrow x\frac{dy}{dx} + y = 0 \) [\( \because xy \neq n\pi \therefore \sin(xy) \neq 0 \)]
\( \Rightarrow \frac{dy}{dx} = \frac{-y}{x} \)

Question. The number of points of discontinuity of \( f \) defined by \( f(x) = |x| - |x + 1| \) is _____________. 
Answer: 0

Question. The function \( f(x) = \frac{2 - x^2}{9x - x^3} \) is discontinuous exactly at _____________ points.
Answer: three

Very Short Answer Questions

Question. If the function \( f \) defined as \( f(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & x \neq 3 \\ k, & x = 3 \end{cases} \) is continuous at \( x = 3 \), find the value of \( k \). 
Answer: It is given that \( f(x) \) is continuous at \( x = 3 \)
\( \therefore \lim_{x \to 3} f(x) = f(3) \)
\( \Rightarrow \lim_{x \to 3} \frac{x^2 - 9}{x - 3} = k \)
\( \Rightarrow \lim_{x \to 3} \frac{(x - 3)(x + 3)}{(x - 3)} = k \)
\( \Rightarrow \lim_{x \to 3} (x + 3) = k \Rightarrow 3 + 3 = k \)
\( \Rightarrow k = 6 \)

Question. For what value of ‘\( k \)’ is the function \( f(x) = \begin{cases} \frac{\sin 5x}{3x} + \cos x, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \) continuous at \( x = 0 \)? 
Answer: \( \lim_{x \to 0^+} f(x) = \lim_{h \to 0} f(0 + h) \)
\( = \lim_{h \to 0} f(h) = \lim_{h \to 0} \left( \frac{\sin 5h}{3h} + \cos h \right) \)
\( = \lim_{h \to 0} \frac{\sin 5h}{5h} \times \frac{5}{3} + \lim_{h \to 0} \cos h = 1 \times \frac{5}{3} + 1 \) [\( \because h \to 0 \Rightarrow 5h \to 0 \)]
\( \Rightarrow \lim_{x \to 0^+} f(x) = \frac{8}{3} \)
Also, \( f(0) = k \)
Since, \( f(x) \) is continuous at \( x = 0 \).
\( \Rightarrow \lim_{x \to 0^+} f(x) = f(0) \Rightarrow \frac{8}{3} = k \)

Question. Determine value of the constant ‘\( k \)’ so that the function \( f(x) = \begin{cases} \frac{kx}{|x|}, & \text{if } x < 0 \\ 3, & \text{if } x \ge 0 \end{cases} \) is continuous at \( x = 0 \). 
Answer: \( \because f(x) \) is continuous at \( x = 0 \)
\( \Rightarrow \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \)
Now, \( \lim_{x \to 0^-} f(x) = \lim_{h \to 0} f(0 - h) \)
\( = \lim_{h \to 0} f(-h) = \lim_{h \to 0} \frac{k(-h)}{|-h|} = \lim_{h \to 0} \frac{-kh}{h} = -k \)
Also, \( f(0) = 3 \)
\( \because \lim_{x \to 0^-} f(x) = f(0) \)
\( \Rightarrow -k = 3 \Rightarrow k = -3 \)

Question. If \( y = 2\sqrt{\sec(e^{2x})} \); then find \( \frac{dy}{dx} \). 
Answer: Given, \( y = 2\sqrt{\sec(e^{2x})} \)
\( \therefore \frac{dy}{dx} = \frac{d}{dx}(2\sqrt{\sec(e^{2x})}) = 2 \times \frac{1}{2\sqrt{\sec(e^{2x})}} \times \sec(e^{2x}) \tan(e^{2x}) \times 2e^{2x} \)
\( = 2\sqrt{\sec(e^{2x})} \tan(e^{2x}) \cdot e^{2x} = 2e^{2x} \sqrt{\sec(e^{2x})} \tan(e^{2x}) \)

Question. If \( y = \text{cosec } (\cot \sqrt{x}) \), then find \( \frac{dy}{dx} \). 
Answer: \( y = \text{cosec}(\cot \sqrt{x}) \)
\( \therefore \frac{dy}{dx} = \frac{d}{dx}(\text{cosec}(\cot \sqrt{x})) \)
\( = -\text{cosec}(\cot \sqrt{x}) \cot(\cot \sqrt{x}) \times (-\text{cosec}^2(\sqrt{x})) \times \frac{1}{2\sqrt{x}} \)
\( = \frac{\text{cosec}(\cot \sqrt{x}) \cot(\cot \sqrt{x}) \text{cosec}^2(\sqrt{x})}{2\sqrt{x}} \)

Question. Find the derivative of \( \log_{10} x \) with respect to \( x \). 
Answer: Let \( y = \log_{10} x = \log_{10} e \cdot \log_e x \)
\( \therefore \frac{dy}{dx} = \log_{10} e \times \frac{1}{x} = \frac{\log_{10} e}{x} \) \( \left[ \because \frac{d}{dx} \log_e x = \frac{1}{x} \right] \)

Question. If \( y = 5e^{7x} + 6e^{-7x} \), show that \( \frac{d^2 y}{dx^2} = 49y \). 
Answer: \( \because y = 5e^{7x} + 6e^{-7x} \Rightarrow \frac{dy}{dx} = 35e^{7x} - 42e^{-7x} \)
\( \Rightarrow \frac{d^2 y}{dx^2} = 245e^{7x} + 294e^{-7x} = 49(5e^{7x} + 6e^{-7x}) = 49y \)

Question. If \( y = \log (\cos e^x) \), then find \( \frac{dy}{dx} \). 
Answer: \( \because y = \log (\cos e^x) \)
\( \Rightarrow \frac{dy}{dx} = \frac{d}{dx} (\log (\cos e^x)) = \frac{1}{\cos(e^x)} \times (-\sin(e^x)) \times e^x \)
\( = \frac{-\sin(e^x) \cdot e^x}{\cos(e^x)} = -e^x \tan(e^x) \)
\( \therefore \frac{dy}{dx} = -e^x \tan(e^x) \)

Short Answer Questions-I

Question. Find \( \frac{dy}{dx} \) at \( t = \frac{2\pi}{3} \) when \( x = 10(t - \sin t) \) and \( y = 12(1 - \cos t) \).
Answer: Given, \( x = 10(t - \sin t) \) and \( y = 12(1 - \cos t) \)
\( \because x = 10(t - \sin t) \Rightarrow \frac{dx}{dt} = 10(1 - \cos t) \) (Differentiating w.r.t. \( t \))
Again \( y = 12(1 - \cos t) \Rightarrow \frac{dy}{dt} = 12(0 + \sin t) = 12 \sin t \) (Differentiating w.r.t. \( t \))
Now, \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{12 \sin t}{10(1 - \cos t)} \)
\( \therefore \left. \frac{dy}{dx} \right]_{t = \frac{2\pi}{3}} = \frac{12 \sin \frac{2\pi}{3}}{10(1 - \cos \frac{2\pi}{3})} = \frac{6}{5} \times \frac{\sin(\pi - \frac{\pi}{3})}{(1 - \cos(\pi - \frac{\pi}{3}))} \)
\( = \frac{6}{5} \times \frac{\sin \frac{\pi}{3}}{1 + \cos \frac{\pi}{3}} = \frac{6}{5} \times \frac{\frac{\sqrt{3}}{2}}{1 + \frac{1}{2}} = \frac{6\sqrt{3}}{5 \times 3} = \frac{2\sqrt{3}}{5} \)

Question. Find \( \frac{dy}{dx} \) at \( x = 1, y = \frac{\pi}{4} \) if \( \sin^2 y + \cos xy = K \). 
Answer: \( \sin^2 y + \cos xy = K \)
Differentiating w.r.t. \( x \), we get
\( 2\sin y \cdot \cos y \frac{dy}{dx} + (-\sin xy)(x \cdot \frac{dy}{dx} + y) = 0 \)
\( \Rightarrow \sin 2y \cdot \frac{dy}{dx} - x \sin xy \frac{dy}{dx} - y \sin xy = 0 \)
\( \Rightarrow \frac{dy}{dx} (\sin 2y - x \sin xy) = y \sin xy \)
\( \Rightarrow \frac{dy}{dx} = \frac{y \sin xy}{(\sin 2y - x \sin xy)} \)
\( \Rightarrow \left. \frac{dy}{dx} \right]_{x=1, y=\frac{\pi}{4}} = \frac{\frac{\pi}{4} \cdot \sin \frac{\pi}{4}}{\sin \frac{\pi}{2} - 1 \cdot \sin \frac{\pi}{4}} = \frac{\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}} = \frac{\frac{\pi}{4\sqrt{2}}}{\frac{\sqrt{2}-1}{\sqrt{2}}} = \frac{\pi}{4(\sqrt{2}-1)} \)

Question. If \( (1 + x)^n = C_0 + C_1x + C_2x^2 + .... + C_nx^n \), then prove that 
(i) \( C_1 + 2C_2 + .... + nC_n = n \cdot 2^{n-1} \)
(ii) \( C_1 - 2C_2 + 3C_3 - .... + (-1)^n nC_n = 0 \)
Answer: We have, \( (1 + x)^n = C_0 + C_1x + C_2x^2 + .... + C_nx^n \)
Differentiating both sides with respect to \( x \), we have
\( n(1 + x)^{n-1} = C_1 + 2C_2x + 3C_3x^2 + .... + nC_nx^{n-1} \)
Putting \( x = 1 \) and \( x = -1 \) successively, we have
(i) \( C_1 + 2C_2 + 3C_3 + .... + nC_n = n \cdot 2^{n-1} \)
and (ii) \( C_1 - 2C_2 + 3C_3 - .... + (-1)^n nC_n = 0 \)

Question. If \( y = (\cos x)^{(\cos x)^{(\cos x) \dots \infty}} \), then show that \( \frac{dy}{dx} = \frac{y^2 \tan x}{y \log \cos x - 1} \).
Answer: We have, \( y = (\cos x)^{(\cos x)^{(\cos x) \dots \infty}} \)
\( \Rightarrow y = (\cos x)^y \)
\( \therefore \log y = \log (\cos x)^y \Rightarrow \log y = y \log \cos x \)
On differentiating w.r.t. \( x \), we get
\( \frac{1}{y} \cdot \frac{dy}{dx} = y \cdot \frac{d}{dx} \log \cos x + \log \cos x \cdot \frac{dy}{dx} \)
\( \Rightarrow \frac{1}{y} \cdot \frac{dy}{dx} = y \cdot \frac{1}{\cos x} \cdot \frac{d}{dx} \cos x + \log \cos x \cdot \frac{dy}{dx} \)
\( \Rightarrow \frac{dy}{dx} \left[ \frac{1}{y} - \log \cos x \right] = \frac{-y \sin x}{\cos x} = -y \tan x \)
\( \therefore \frac{dy}{dx} = \frac{-y^2 \tan x}{(1 - y \log \cos x)} = \frac{y^2 \tan x}{y \log \cos x - 1} \)

Question. If \( x = a \cos \theta ; y = b \sin \theta \), then find \( \frac{d^2 y}{dx^2} \). 
Answer: We have,
\( x = a \cos \theta \Rightarrow \frac{dx}{d\theta} = -a \sin \theta \)
and \( y = b \sin \theta \Rightarrow \frac{dy}{d\theta} = b \cos \theta \)
\( \therefore \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{b \cos \theta}{-a \sin \theta} = -\frac{b}{a} \cot \theta \)
Again diff. w.r.t \( x \), we have
\( \frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{b}{a} \frac{d}{dx} (-\cot \theta) = -\frac{b}{a} \times (-\text{cosec}^2 \theta) \frac{d\theta}{dx} \)
\( = \frac{b}{a} \times (\text{cosec}^2 \theta) \times \left( -\frac{1}{a \sin \theta} \right) \)
\( \therefore \frac{d^2 y}{dx^2} = -\frac{b}{a^2} \text{cosec}^3 \theta \)

Question. If \( y^x = e^{y - x} \), then prove that \( \frac{dy}{dx} = \frac{(1 + \log y)^2}{\log y} \). 
Answer: Given, \( y^x = e^{y - x} \)
Taking logarithm both sides, we get \( \log y^x = \log e^{y - x} \)
\( \Rightarrow x. \log y = (y - x) . \log e \Rightarrow x . \log y = y - x \)
\( \Rightarrow x(1 + \log y) = y \Rightarrow x = \frac{y}{1 + \log y} \)
Differentiating both sides with respect to \( y \), we get
\( \frac{dx}{dy} = \frac{(1 + \log y).1 - y.\left(0 + \frac{1}{y}\right)}{(1 + \log y)^2} = \frac{1 + \log y - 1}{(1 + \log y)^2} = \frac{\log y}{(1 + \log y)^2} \)
\( \Rightarrow \frac{dy}{dx} = \frac{(1 + \log y)^2}{\log y} \)
[\( \because \) (i) \( \log_e mn = \log_e m + \log_e n \), (ii) \( \log_e \frac{m}{n} = \log_e m - \log_e n \), (iii) \( \log_e m^n = n \log_e m \), (iv) \( \log_e e = 1 \)]

Question. If \( (\cos x)^y = (\cos y)^x \), then find \( \frac{dy}{dx} \). 
Answer: Given, \( (\cos x)^y = (\cos y)^x \)
Taking logarithm both sides, we get \( \log(\cos x)^y = \log(\cos y)^x \)
\( \Rightarrow y . \log(\cos x) = x . \log(\cos y) \) [\( \because \log m^n = n \log m \)]
Differentiating both sides with respect to \( x \), we get
\( y \cdot \frac{1}{\cos x} (-\sin x) + \log(\cos x) \cdot \frac{dy}{dx} = x \cdot \frac{1}{\cos y} \cdot (-\sin y) \cdot \frac{dy}{dx} + \log(\cos y) \)
\( \Rightarrow -\frac{y \sin x}{\cos x} + \log(\cos x) \cdot \frac{dy}{dx} = -\frac{x \sin y}{\cos y} \cdot \frac{dy}{dx} + \log(\cos y) \)
\( \Rightarrow \log(\cos x) \cdot \frac{dy}{dx} + \frac{x \sin y}{\cos y} \cdot \frac{dy}{dx} = \log(\cos y) + \frac{y \sin x}{\cos x} \)
\( \Rightarrow \frac{dy}{dx} \left[ \log(\cos x) + \frac{x \sin y}{\cos y} \right] = \log(\cos y) + \frac{y \sin x}{\cos x} \)
\( \Rightarrow \frac{dy}{dx} = \frac{\log(\cos y) + \frac{y \sin x}{\cos x}}{\log(\cos x) + \frac{x \sin y}{\cos y}} = \frac{\log(\cos y) + y \tan x}{\log(\cos x) + x \tan y} \)

Question. Find the value of \( \frac{dy}{dx} \) at \( \theta = \frac{\pi}{4} \), if \( x = ae^{\theta}(\sin \theta - \cos \theta) \) and \( y = ae^{\theta}(\sin \theta + \cos \theta) \). 
Answer: Given, \( x = ae^{\theta}(\sin \theta - \cos \theta) \) and \( y = ae^{\theta}(\sin \theta + \cos \theta) \)
Taking \( x = ae^{\theta}(\sin \theta - \cos \theta) \)
Differentiating with respect to \( \theta \), we get
\( \frac{dx}{d\theta} = ae^{\theta}(\cos \theta + \sin \theta) + a(\sin \theta - \cos \theta).e^{\theta} = ae^{\theta}(\cos \theta + \sin \theta + \sin \theta - \cos \theta) \)
\( = 2ae^{\theta} \sin \theta \) ...(i)
Again, \( y = ae^{\theta}(\sin \theta + \cos \theta) \)
Differentiating with respect to \( \theta \), we get
\( \frac{dy}{d\theta} = ae^{\theta}(\cos \theta - \sin \theta) + a(\sin \theta + \cos \theta).e^{\theta} = ae^{\theta}(\cos \theta - \sin \theta + \sin \theta + \cos \theta) \)
\( = 2ae^{\theta} \cos \theta \) ...(ii)
\( \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2ae^{\theta} \cos \theta}{2ae^{\theta} \sin \theta} = \cot \theta \) [From (i) and (ii)]
\( \Rightarrow \frac{dy}{dx} = \cot \theta \Rightarrow \left[ \frac{dy}{dx} \right]_{\theta = \frac{\pi}{4}} = \cot \frac{\pi}{4} = 1 \)

Question. Differentiate the following with respect to \( x \): \( (\sin x)^x + (\cos x)^{\sin x} \). 
Answer: Let \( u = (\sin x)^x \) and \( v = (\cos x)^{\sin x} \)
\( \therefore \) Given differential equation becomes \( y = u + v \)
\( \Rightarrow \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ....(i)
Now, \( u = (\sin x)^x \)
Taking log on both sides, we get \( \log u = x \log \sin x \)
Differentiating with respect to \( x \), we get
\( \frac{1}{u} \cdot \frac{du}{dx} = x \cdot \frac{1}{\sin x} \cdot \cos x + \log \sin x \Rightarrow \frac{du}{dx} = u(x \cot x + \log \sin x) \)
\( \Rightarrow \frac{du}{dx} = (\sin x)^x [x \cot x + \log \sin x] \) ....(ii)
Again \( v = (\cos x)^{\sin x} \)
Taking log on both sides, we get \( \log v = \sin x \cdot \log \cos x \)
Differentiating with respect to \( x \), we get
\( \frac{1}{v} \cdot \frac{dv}{dx} = \sin x \cdot \frac{1}{\cos x} (-\sin x) + \log(\cos x) \cdot \cos x \)
\( \Rightarrow \frac{dv}{dx} = v \left[ -\frac{\sin^2 x}{\cos x} + \cos x \cdot \log \cos x \right] = (\cos x)^{\sin x} [\cos x \cdot \log(\cos x) - \frac{\sin^2 x}{\cos x}] \)
\( = (\cos x)^{1+\sin x} [\log(\cos x) - \tan^2 x] \) (Simplification variant dependent on steps) ....(iii)
From (i), (ii) and (iii), we get
\( \frac{dy}{dx} = (\sin x)^x [x \cot x + \log \sin x] + (\cos x)^{\sin x} [\cos x \log(\cos x) - \sin x \tan x] \)

Question. If \( \cos y = x \cos(a + y) \), with \( \cos a \neq \pm 1 \), then prove that \( \frac{dy}{dx} = \frac{\cos^2(a + y)}{\sin a} \). Hence show that \( \sin a \frac{d^2y}{dx^2} + \sin 2(a + y) \frac{dy}{dx} = 0 \). 
Answer: Given, \( \cos y = x \cos(a + y) \)
\( \therefore x = \frac{\cos y}{\cos(a + y)} \)
Differentiating with respect to \( y \) on both sides, we get
\( \frac{dx}{dy} = \frac{\cos(a + y) \times (-\sin y) - \cos y \times [-\sin(a + y)]}{\cos^2(a + y)} \)
\( \Rightarrow \frac{dx}{dy} = \frac{\cos y \sin(a + y) - \sin y \cos(a + y)}{\cos^2(a + y)} \Rightarrow \frac{dx}{dy} = \frac{\sin(a + y - y)}{\cos^2(a + y)} \)
\( \Rightarrow \frac{dx}{dy} = \frac{\sin a}{\cos^2(a + y)} \)
\( \therefore \frac{dy}{dx} = \frac{\cos^2(a + y)}{\sin a} \)
Differentiating both sides w.r.t. \( x \), we get
\( \frac{d^2 y}{dx^2} = \frac{1}{\sin a} \left[ 2 \cos(a + y) \cdot (-\sin(a + y)) \cdot \frac{dy}{dx} \right] \)
\( \Rightarrow \sin a \frac{d^2 y}{dx^2} = - \sin 2(a + y) \cdot \frac{dy}{dx} \Rightarrow \sin a \frac{d^2 y}{dx^2} + \sin 2(a + y) \frac{dy}{dx} = 0 \)

Question. If \( x = a \sin 2t (1 + \cos 2t) \) and \( y = b \cos 2t (1 - \cos 2t) \), then show that \( \frac{dy}{dx} \) at \( t = \frac{\pi}{4} = \frac{b}{a} \). Also find the value of \( \frac{dy}{dx} \) at \( t = \frac{\pi}{3} \). 
Answer: Given, \( x = a \sin 2t (1 + \cos 2t) \) and \( y = b \cos 2t (1 - \cos 2t) \)
\( \Rightarrow \frac{dx}{dt} = a [\sin 2t \times (-2 \sin 2t) + (1 + \cos 2t) \times 2 \cos 2t] = a [-2 \sin^2 2t + 2 \cos 2t + 2 \cos^2 2t] \)
\( = a(2 \cos 4t + 2 \cos 2t) = 2a (\cos 4t + \cos 2t) \)
Again, \( \frac{dy}{dt} = b [\cos 2t \times 2 \sin 2t + (1 - \cos 2t) \times (-2 \sin 2t)] \)
\( = b [\sin 4t - 2 \sin 2t + \sin 4t] = b [2 \sin 4t - 2 \sin 2t] = 2b (\sin 4t - \sin 2t) \)
\( \therefore \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2b(\sin 4t - \sin 2t)}{2a(\cos 4t + \cos 2t)} = \frac{b(\sin 4t - \sin 2t)}{a(\cos 4t + \cos 2t)} \)
So, \( \left( \frac{dy}{dx} \right)_{t = \frac{\pi}{4}} = \frac{b}{a} \left[ \frac{\sin \pi - \sin \frac{\pi}{2}}{\cos \pi + \cos \frac{\pi}{2}} \right] = \frac{b}{a} \times \left( \frac{0 - 1}{-1 + 0} \right) = \frac{b}{a} \) and
\( \left( \frac{dy}{dx} \right)_{t = \frac{\pi}{3}} = \frac{b}{a} \left[ \frac{\sin \frac{4\pi}{3} - \sin \frac{2\pi}{3}}{\cos \frac{4\pi}{3} + \cos \frac{2\pi}{3}} \right] = \frac{b}{a} \left[ \frac{-\sin \frac{\pi}{3} - \sin \frac{\pi}{3}}{-\cos \frac{\pi}{3} - \cos \frac{\pi}{3}} \right] \)
\( = \frac{b}{a} \times \left[ \frac{-2 \sin \frac{\pi}{3}}{-2 \cos \frac{\pi}{3}} \right] = \frac{b}{a} \tan \frac{\pi}{3} = \frac{\sqrt{3}b}{a} \)

Question. If \( \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \), then show that \( \frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}} \). 
Answer: Given, \( \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \)
Putting \( x = \sin \alpha \Rightarrow \alpha = \sin^{-1} x \) and \( y = \sin \beta \Rightarrow \beta = \sin^{-1} y \), we get
\( \sqrt{1 - \sin^2 \alpha} + \sqrt{1 - \sin^2 \beta} = a(\sin \alpha - \sin \beta) \Rightarrow \cos \alpha + \cos \beta = a(\sin \alpha - \sin \beta) \)
\( \Rightarrow 2 \cos \frac{(\alpha + \beta)}{2} \cos \frac{(\alpha - \beta)}{2} = a . 2 \cos \left( \frac{\alpha + \beta}{2} \right) \sin \left( \frac{\alpha - \beta}{2} \right) \)
\( \Rightarrow \cot \left( \frac{\alpha - \beta}{2} \right) = a \Rightarrow \frac{\alpha - \beta}{2} = \cot^{-1} a \Rightarrow \alpha - \beta = 2 \cot^{-1} a \)
\( \Rightarrow \sin^{-1} x - \sin^{-1} y = 2 \cot^{-1} a \)
Differentiating both sides with respect to \( x \), we get
\( \frac{1}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \sqrt{\frac{1 - y^2}{1 - x^2}} \)

Question. If \( x = \cos t(3 - 2 \cos^2 t) \) and \( y = \sin t (3 - 2 \sin^2 t) \), then find the value of \( \frac{dy}{dx} \) at \( t = \frac{\pi}{4} \). 
Answer: Given, \( x = \cos t(3 - 2 \cos^2 t) \)
Differentiating both sides with respect to \( t \), we get
\( \frac{dx}{dt} = \cos t [0 + 4 \cos t \cdot \sin t] + (3 - 2 \cos^2 t).(-\sin t) \)
\( = 4 \sin t \cdot \cos^2 t - 3 \sin t + 2 \cos^2 t \cdot \sin t \)
\( = 6 \sin t \cos^2 t - 3 \sin t = 3 \sin t (2 \cos^2 t - 1) = 3 \sin t \cdot \cos 2t \)
Again, \( y = \sin t (3 - 2 \sin^2 t) \)
Differentiating both sides with respect to \( t \), we get
\( \frac{dy}{dt} = \sin t . [0 - 4 \sin t \cos t] + (3 - 2 \sin^2 t) . \cos t \)
\( = - 4 \sin^2 t \cdot \cos t + 3 \cos t - 2 \sin^2 t \cdot \cos t = 3 \cos t - 6 \sin^2 t \cdot \cos t \)
\( = 3 \cos t (1 - 2 \sin^2 t) = 3 \cos t \cdot \cos 2t \)
Now, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3 \cos t \cdot \cos 2t}{3 \sin t \cdot \cos 2t} \Rightarrow \frac{dy}{dx} = \cot t \)
\( \therefore \left. \frac{dy}{dx} \right]_{t = \frac{\pi}{4}} = \cot \frac{\pi}{4} = 1 \)

Question. Differentiate \( \tan^{-1} \left( \frac{\sqrt{1 - x^2}}{x} \right) \) with respect to \( \cos^{-1}(2x \sqrt{1 - x^2}) \), when \( x \neq 0 \). 
Answer: Let \( u = \tan^{-1} \left( \frac{\sqrt{1 - x^2}}{x} \right) \) and \( v = \cos^{-1}(2x \sqrt{1 - x^2}) \)
We have to determine \( \frac{du}{dv} \)
Put \( x = \sin \theta \Rightarrow \theta = \sin^{-1} x \)
Now, \( u = \tan^{-1} \left( \frac{\sqrt{1 - \sin^2 \theta}}{\sin \theta} \right) \Rightarrow u = \tan^{-1} \left( \frac{\cos \theta}{\sin \theta} \right) \)
\( \Rightarrow u = \tan^{-1}(\cot \theta) \Rightarrow u = \tan^{-1} \left[ \tan \left( \frac{\pi}{2} - \theta \right) \right] \)
\( \Rightarrow u = \frac{\pi}{2} - \theta \Rightarrow u = \frac{\pi}{2} - \sin^{-1} x \)
\( \Rightarrow \frac{du}{dx} = 0 - \frac{1}{\sqrt{1 - x^2}} \Rightarrow \frac{du}{dx} = -\frac{1}{\sqrt{1 - x^2}} \)
Again, \( v = \cos^{-1}(2x \sqrt{1 - x^2}) \)
\( \because x = \sin \theta \)
\( \therefore v = \cos^{-1}(2 \sin \theta \sqrt{1 - \sin^2 \theta}) \Rightarrow v = \cos^{-1}(2 \sin \theta \cdot \cos \theta) \)
\( \Rightarrow v = \cos^{-1}(\sin 2\theta) \Rightarrow v = \cos^{-1} \left[ \cos \left( \frac{\pi}{2} - 2\theta \right) \right] \)
\( \Rightarrow v = \frac{\pi}{2} - 2\theta \Rightarrow v = \frac{\pi}{2} - 2 \sin^{-1} x \)
\( \Rightarrow \frac{dv}{dx} = 0 - \frac{2}{\sqrt{1 - x^2}} \Rightarrow \frac{dv}{dx} = -\frac{2}{\sqrt{1 - x^2}} \)
\( \therefore \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{-\frac{1}{\sqrt{1 - x^2}}}{-\frac{2}{\sqrt{1 - x^2}}} = \frac{1}{2} \)

Question. Differentiate the following function with respect to \( x \): \( y = (\sin x)^x + \sin^{-1} \sqrt{x} \). 
Answer: Given, \( y = (\sin x)^x + \sin^{-1} \sqrt{x} \)
\( y = u + v \), where \( u = (\sin x)^x \), \( v = \sin^{-1} \sqrt{x} \)
\( \therefore \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \) ...(i)
Now, \( u = (\sin x)^x \)
Taking log both sides, we get \( \log u = \log (\sin x)^x \Rightarrow \log u = x . \log(\sin x) \)
Differentiating both sides with respect to \( x \), we get
\( \frac{1}{u} \cdot \frac{du}{dx} = x \cdot \frac{1}{\sin x} \cos x + \log \sin x \Rightarrow \frac{du}{dx} = u(x \cot x + \log \sin x) \)
\( \Rightarrow \frac{du}{dx} = (\sin x)^x (x \cot x + \log \sin x) \) ...(ii)
Also, \( v = \sin^{-1} \sqrt{x} \)
\( \frac{dv}{dx} = \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \times \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x} \sqrt{1 - x}} \) ...(iii)
From (i), (ii) and (iii), we get
\( \therefore \frac{dy}{dx} = (\sin x)^x (x \cot x + \log \sin x) + \frac{1}{2\sqrt{x(1 - x)}} \)

Question. If \( x^{13}y^7 = (x + y)^{20} \), then prove that \( \frac{dy}{dx} = \frac{y}{x} \). 
OR
If \( x^m y^n = (x + y)^{m+n} \), then prove that \( \frac{dy}{dx} = \frac{y}{x} \). 
Answer: Given \( x^{13}y^7 = (x + y)^{20} \)
Taking logarithm on both sides, we get \( \log(x^{13}y^7) = \log(x + y)^{20} \)
\( \Rightarrow \log x^{13} + \log y^7 = 20 \log(x + y) \Rightarrow 13 \log x + 7 \log y = 20 \log(x + y) \)
Differentiating both sides with respect to \( x \), we get
\( \frac{13}{x} + \frac{7}{y} \cdot \frac{dy}{dx} = \frac{20}{x + y} \left( 1 + \frac{dy}{dx} \right) \Rightarrow \frac{13}{x} - \frac{20}{x + y} = \left( \frac{20}{x + y} - \frac{7}{y} \right) \frac{dy}{dx} \)
\( \Rightarrow \frac{13x + 13y - 20x}{x(x + y)} = \left( \frac{20y - 7x - 7y}{(x + y)y} \right) \frac{dy}{dx} \Rightarrow \frac{13y - 7x}{x(x + y)} = \left( \frac{13y - 7x}{y(x + y)} \right) \cdot \frac{dy}{dx} \)
\( \Rightarrow \frac{dy}{dx} = \frac{13y - 7x}{x(x + y)} \times \frac{y(x + y)}{13y - 7x} \Rightarrow \frac{dy}{dx} = \frac{y}{x} \)
OR
Do yourself (similar question)

Question. Differentiate with respect to \( x \): \( \sin^{-1} \left( \frac{2^{x+1} \cdot 3^x}{1 + (36)^x} \right) \). 
Answer: Let \( y = \sin^{-1} \left( \frac{2^{x+1} \cdot 3^x}{1 + (36)^x} \right) = \sin^{-1} \left( \frac{2 \cdot 2^x \cdot 3^x}{1 + (6^2)^x} \right) = \sin^{-1} \left( \frac{2 \cdot 6^x}{1 + (6^x)^2} \right) \)
Let \( 6^x = \tan \theta \Rightarrow \theta = \tan^{-1} (6^x) \)
\( \therefore y = \sin^{-1} \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right) \Rightarrow y = \sin^{-1} (\sin 2\theta) \)
\( \Rightarrow y = 2\theta \Rightarrow y = 2 . \tan^{-1} (6^x) \)
\( \Rightarrow \frac{dy}{dx} = \frac{2}{1 + (6^x)^2} \cdot 6^x \cdot \log_e 6 \Rightarrow \frac{dy}{dx} = \frac{2 \cdot 6^x \cdot \log_e 6}{1 + 36^x} \)

Question. If \( x \sin(a + y) + \sin a \cos(a + y) = 0 \), then prove that \( \frac{dy}{dx} = \frac{\sin^2(a + y)}{\sin a} \). 
Answer: Given \( x \sin(a + y) + \sin a \cos(a + y) = 0 \)
\( \Rightarrow x = -\frac{\sin a \cos(a + y)}{\sin(a + y)} \Rightarrow x = - \sin a . \cot(a + y) \)
Differentiating with respect to \( y \), we get
\( \frac{dx}{dy} = + \sin a . \text{cosec}^2(a + y) = \frac{\sin a}{\sin^2(a + y)} \)
\( \Rightarrow \frac{dy}{dx} = \frac{\sin^2(a + y)}{\sin a} \)

Question. If \( e^x + e^y = e^{x + y} \), then prove that \( \frac{dy}{dx} + e^{y - x} = 0 \). 
Answer: Given, \( e^x + e^y = e^{x + y} \)
Differentiating both sides with respect to \( x \), we get
\( e^x + e^y \cdot \frac{dy}{dx} = e^{x + y} \left\{ 1 + \frac{dy}{dx} \right\} \)
\( \Rightarrow e^x + e^y \cdot \frac{dy}{dx} = e^{x + y} + e^{x + y} \cdot \frac{dy}{dx} \Rightarrow (e^y - e^{x + y}) \frac{dy}{dx} = e^{x + y} - e^x \)
\( \Rightarrow (e^y - e^x - e^y) \frac{dy}{dx} = e^x + e^y - e^x \) [\( \because e^x + e^y = e^{x + y} \) (given)]
\( \Rightarrow e^x \cdot \frac{dy}{dx} = - e^y \Rightarrow \frac{dy}{dx} = - \frac{e^y}{e^x} \)
\( \Rightarrow \frac{dy}{dx} = - e^{y - x} \Rightarrow \frac{dy}{dx} + e^{y - x} = 0 \)

Question. If \( x = e^{\cos 2t} \) and \( y = e^{\sin 2t} \), prove that \( \frac{dy}{dx} = - \frac{y \log x}{x \log y} \).
Answer: We have \( x = e^{\cos 2t} \)
\( \frac{dx}{dt} = e^{\cos 2t} (-2 \sin 2t) = -2x \sin 2t \) [Differentiating w.r.t. \( t \)]
Again \( y = e^{\sin 2t} \)
\( \frac{dy}{dt} = e^{\sin 2t} . 2 \cos 2t = 2y \cos 2t \) [Differentiating w.r.t. \( t \)]
Now, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2y \cos 2t}{-2x \sin 2t} \Rightarrow \frac{dy}{dx} = - \frac{y \cos 2t}{x \sin 2t} \)
\( \Rightarrow \frac{dy}{dx} = - \frac{y \log x}{x \log y} \) [\( \because x = e^{\cos 2t} \Rightarrow \log x = \cos 2t; y = e^{\sin 2t} \Rightarrow \log y = \sin 2t \)]
Hence proved.

Question. If \( x \in R - [-1, 1] \) then prove that the derivative of \( \sec^{-1}x \) with respect to \( x \) is \( \frac{1}{|x| \sqrt{x^2 - 1}} \). 
Answer: Let \( y = \sec^{-1} x \)
Then, \( \sec y = \sec(\sec^{-1} x) = x \)
Differentiating both sides with respect to \( x \), we have
\( \frac{d}{dx} \sec y = \frac{d}{dx} (x) \Rightarrow \frac{d}{dy}(\sec y) \frac{dy}{dx} = 1 \)
\( \Rightarrow \sec y \tan y \frac{dy}{dx} = 1 \) [Using chain rule]
\( \Rightarrow \frac{dy}{dx} = \frac{1}{\sec y \tan y} = \frac{1}{|\sec y| . |\tan y|} \)
\( \Rightarrow \frac{dy}{dx} = \frac{1}{|\sec y| \sqrt{\tan^2 y}} = \frac{1}{|\sec y| \sqrt{\sec^2 y - 1}} = \frac{1}{|x| \sqrt{x^2 - 1}} \)
[Note: If \( x > 1 \), then \( y \in (0, \frac{\pi}{2}) \) \( \therefore \sec y > 0, \tan y > 0 \Rightarrow |\sec y| . |\tan y| = \sec y \tan y \). If \( x < -1 \), then \( y \in (\frac{\pi}{2}, \pi) \) \( \therefore \sec y < 0, \tan y < 0 \Rightarrow |\sec y| . |\tan y| = (-\sec y)(-\tan y) = \sec y \tan y \)]

SECOND ORDER DERIVATIVES

Question. If \( x = a \cos \theta + b \sin \theta \) and \( y = a \sin \theta - b \cos \theta \), then show that \( y^2 \frac{d^2y}{dx^2} - x \frac{dy}{dx} + y = 0 \).
Answer: Given, \( x = a \cos \theta + b \sin \theta \Rightarrow \frac{dx}{d\theta} = -a \sin \theta + b \cos \theta \) ...(i)
Also, \( y = a \sin \theta - b \cos \theta \Rightarrow \frac{dy}{d\theta} = a \cos \theta + b \sin \theta \) ...(ii)
\( \therefore \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{a \cos \theta + b \sin \theta}{-a \sin \theta + b \cos \theta} \) [From (i) and (ii)]
\( \Rightarrow \frac{dy}{dx} = \frac{a \cos \theta + b \sin \theta}{b \cos \theta - a \sin \theta} \Rightarrow \frac{dy}{dx} = \frac{x}{y} \) (Note: based on \( x, y \) definitions) ...(iii)
Differentiating again with respect to \( x \), we get
\( \frac{d^2 y}{dx^2} = - \frac{y - x \cdot \frac{dy}{dx}}{y^2} \Rightarrow y^2 \frac{d^2 y}{dx^2} = -y + x \frac{dy}{dx} \Rightarrow y^2 \frac{d^2 y}{dx^2} - x \frac{dy}{dx} + y = 0 \)

Question. If \( y = Pe^{ax} + Qe^{bx} \), then show that \( \frac{d^2y}{dx^2} - (a+b) \frac{dy}{dx} + aby = 0 \). 
Answer: Given, \( y = Pe^{ax} + Qe^{bx} \)
On differentiating with respect to \( x \), we have \( \frac{dy}{dx} = Pae^{ax} + Qbe^{bx} \)
Again, differentiating with respect to \( x \), we have \( \frac{d^2 y}{dx^2} = Pa^2e^{ax} + Qb^2e^{bx} \)
Now, LHS \( = \frac{d^2 y}{dx^2} - (a + b) \frac{dy}{dx} + aby \)
\( = Pa^2e^{ax} + Qb^2e^{bx} - (a + b)(Pae^{ax} + Qbe^{bx}) + ab (Pe^{ax} + Qe^{bx}) \)
\( = Pa^2e^{ax} + Qb^2e^{bx} - Pa^2e^{ax} - Pabe^{ax} - Qabebx - Qb^2e^{bx} + Pabe^{ax} + Qabebx \)
\( = 0 = RHS \)

Question. If \( y = e^{a \cos^{-1} x} \), \( -1 < x < 1 \), then show that \( (1 - x^2) \frac{d^2y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0 \). 
Answer: Given, \( y = e^{a \cos^{-1} x} \), \( -1 < x < 1 \)
Differentiating w.r.t \( x \), we have
\( \therefore \frac{dy}{dx} = e^{a \cos^{-1} x} \times a \times \frac{-1}{\sqrt{1 - x^2}} = - \frac{ay}{\sqrt{1 - x^2}} \)
\( \Rightarrow \sqrt{1 - x^2} \frac{dy}{dx} = - ay \)
Squaring both sides, we have \( (1 - x^2) \left( \frac{dy}{dx} \right)^2 = a^2 y^2 \)
Again differentiating w.r.t. \( x \), we have
\( (1 - x^2) \times 2 \frac{dy}{dx} \cdot \frac{d^2 y}{dx^2} + \left( \frac{dy}{dx} \right)^2 \times (-2x) = a^2 \times 2y \frac{dy}{dx} \)
\( \Rightarrow (1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = a^2 y \Rightarrow (1 - x^2) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} - a^2 y = 0 \) Proved

Question. If \( y = \sin(\log x) \), then prove that \( x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0 \). 
Answer: Given, \( y = \sin(\log x) \)
\( \Rightarrow \frac{dy}{dx} = \cos(\log x) \times \frac{1}{x} = \frac{\cos(\log x)}{x} \)
Again, \( \frac{d^2 y}{dx^2} = \frac{x \left[ -\sin(\log x) \times \frac{1}{x} \right] - \cos(\log x) \cdot 1}{x^2} = \frac{-\sin(\log x) - \cos(\log x)}{x^2} \)
Now, LHS \( = x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + y \)
\( = x^2 \frac{\{-\sin(\log x) - \cos(\log x)\}}{x^2} + x \frac{\cos(\log x)}{x} + \sin(\log x) \)
\( = - \cos(\log x) - \sin(\log x) + \cos(\log x) + \sin(\log x) = 0 = RHS \)

Question. If \( y = \text{cosec}^{-1} x \), \( x > 1 \), then show that \( x(x^2 - 1) \frac{d^2y}{dx^2} + (2x^2 - 1) \frac{dy}{dx} = 0 \). 
Answer: \( \because y = \text{cosec}^{-1} x \)
Differentiating with respect to \( x \), we get \( \therefore \frac{dy}{dx} = - \frac{1}{x \sqrt{x^2 - 1}} \)
Again differentiating with respect to \( x \), we get
\( \frac{d^2 y}{dx^2} = - \frac{x \sqrt{x^2 - 1} \cdot 0 + 1 \cdot \left[ x \cdot \frac{2x}{2 \sqrt{x^2 - 1}} + \sqrt{x^2 - 1} \right]}{x^2(x^2 - 1)} \)
\( \Rightarrow \frac{d^2 y}{dx^2} = \frac{\frac{x^2 + x^2 - 1}{\sqrt{x^2 - 1}}}{x^2(x^2 - 1)} = \frac{2x^2 - 1}{\sqrt{x^2 - 1} \cdot x^2 (x^2 - 1)} \)
\( \Rightarrow x(x^2 - 1) \frac{d^2 y}{dx^2} = \frac{2x^2 - 1}{x \sqrt{x^2 - 1}} = (2x^2 - 1) \left( - \frac{dy}{dx} \right) \)
\( \Rightarrow x(x^2 - 1) \frac{d^2 y}{dx^2} + (2x^2 - 1) \frac{dy}{dx} = 0 \)

Question. If \( y = 3 \cos(\log x) + 4 \sin(\log x) \), show that \( x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0 \). 
Answer: Given, \( y = 3 \cos(\log x) + 4 \sin(\log x) \)
Differentiating with respect to \( x \), we get \( \frac{dy}{dx} = \frac{-3 \sin(\log x)}{x} + \frac{4 \cos(\log x)}{x} \)
\( \Rightarrow y_1 = \frac{1}{x} [-3 \sin(\log x) + 4 \cos(\log x)] \)
Again differentiating with respect to \( x \), we get
\( \frac{d^2 y}{dx^2} = \frac{x \left[ \frac{-3 \cos(\log x)}{x} - \frac{4 \sin(\log x)}{x} \right] - [-3 \sin(\log x) + 4 \cos(\log x)]}{x^2} \)
\( = \frac{-3 \cos(\log x) - 4 \sin(\log x) + 3 \sin(\log x) - 4 \cos(\log x)}{x^2} \)
\( \frac{d^2 y}{dx^2} = \frac{- \sin(\log x) - 7 \cos(\log x)}{x^2} \Rightarrow y_2 = \frac{- \sin(\log x) - 7 \cos(\log x)}{x^2} \)
Now, LHS \( = x^2 y_2 + xy_1 + y \)
\( = x^2 \left( \frac{- \sin(\log x) - 7 \cos(\log x)}{x^2} \right) + x \times \frac{1}{x} [-3 \sin(\log x) + 4 \cos(\log x)] + 3 \cos(\log x) + 4 \sin(\log x) \)
\( = - \sin(\log x) - 7 \cos(\log x) - 3 \sin(\log x) + 4 \cos(\log x) + 3 \cos(\log x) + 4 \sin(\log x) \)
\( = 0 = RHS \)

Question. If \( x = a(\cos t + t \sin t) \) and \( y = a(\sin t - t \cos t) \), \( 0 < t < \frac{\pi}{2} \), find \( \frac{d^2x}{dt^2} \), \( \frac{d^2y}{dt^2} \) and \( \frac{d^2y}{dx^2} \).
Answer: Given, \( x = a(\cos t + t \sin t) \)
Differentiating both sides with respect to \( t \), we get \( \frac{dx}{dt} = a(-\sin t + t \cos t + \sin t) \Rightarrow \frac{dx}{dt} = at \cos t \) ...(i)
Differentiating again with respect to \( t \), we get \( \frac{d^2 x}{dt^2} = a(-t \sin t + \cos t) = a(\cos t - t \sin t) \)
Again, \( y = a(\sin t - t \cos t) \)
Differentiating with respect to \( t \), we get \( \frac{dy}{dt} = a(\cos t + t \sin t - \cos t) \Rightarrow \frac{dy}{dt} = at \sin t \) ...(ii)
Differentiating again with respect to \( t \) we get \( \frac{d^2 y}{dt^2} = a(t \cos t + \sin t) \)
Now, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) [From (i) and (ii)]
\( \Rightarrow \frac{dy}{dx} = \frac{at \sin t}{at \cos t} \Rightarrow \frac{dy}{dx} = \tan t \)
Differentiating again with respect to \( x \), we get
\( \frac{d^2 y}{dx^2} = \sec^2 t \cdot \frac{dt}{dx} = \sec^2 t \cdot \frac{1}{dx/dt} = \frac{\sec^2 t}{at \cos t} = \frac{\sec^3 t}{at} \) [From (i)]
Hence, \( \frac{d^2 x}{dt^2} = a(\cos t - t \sin t) \), \( \frac{d^2 y}{dt^2} = a(t \cos t + \sin t) \) and \( \frac{d^2 y}{dx^2} = \frac{\sec^3 t}{at} \).