Miscellaneous Exercise Class 12 Chapter 5

Let \( y=\left(3 x^{2}-9 x+5\right)^{9} \)
If \( \mathrm{u}=\mathrm{v}(\mathrm{w}(x)) \)
Then using chain rule \( \frac{d u}{d x}=\frac{d v}{d w} \times \frac{d w}{d x} \)
\( \therefore \) Differentiating y w.r.t. \(x\) using chain rule
\(\frac{d y}{d x}=\frac{d}{d x}\left(3 x^{2}-9 x+5\right)^{9}\)
\(=9\left(3 x^{2}-9 x+5\right)^{8} \times \frac{d}{d x}(3 x^2-9 x+5)\)
\(=9\left(3 x^{2}-9 x+5\right)^{8} \times(6 x-9)\)
\(=9\left(3 x^{2}-9 x+5\right)^{8} \times 3(2 x-3)\)
\(=27\left(3 x^{2}-9 x+5\right)^{8}(2 x-3)\)
\(\therefore \frac{d y}{d x}=27\left(3 x^{2}-9 x+5\right)^{8}(2 x-3)\)

Let \( y=\sin ^{3} x+\cos ^{6} x \)
Differentiating both sides with respect to \(x\)
\(\therefore \frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{3} x\right)+\frac{d}{d x}\left(\cos ^{6} x\right) \because \frac{d}{d x}(\sin x)=\cos x \& \frac{d}{d x}(\cos x)=-\sin x\)
\(=3 \sin^2 x \times \frac{d}{d x}(\sin x)+6 \cos^5 x \times \frac{d}{d x}(\cos x)\)
\(=3 \sin ^{2} x \times \cos x+6 \cos ^{5} x \times(-\sin x)\)
\(=3 \sin x \cos x\left(\sin x-2 \cos ^{4} x\right)\)
\(\therefore \frac{d y}{d x}=3 \sin x \cos x\left(\sin x-2 \cos ^{4} x\right)\)

Let \( y=(5 x)^{3 \cos 2 x} \)
Then \( \log y=\log (5 x)^{3 \cos 2 x} \)
\(\Rightarrow \log y=3 \cos^{2 x} \times \log 5 x\)
Differentiating both sides with respect to \( x \), we get
\(\frac{1}{y} \frac{d y}{d x}=3\left[\log 5 x \times \frac{d}{d x}(\cos 2 x)+\cos 2 x \times \frac{d}{d x}(\log 5 x)\right]\)
\({\left[\therefore \frac{d}{d x}(u v)=u \times \frac{d v}{d x}+v \times \frac{d u}{d x}\right]}\)
\(=\frac{d y}{d x}=3 y\left[\log 5 x(-2 \sin 2 x) \times \frac{d}{d x}(2 x)+\cos 2 x \times \frac{1}{5 x} \times \frac{d}{d x}(5 x)\right]\)
\(=\frac{d y}{d x}=3 y[-2 \sin 2 x \log 5 x+] \frac{\cos 2 x}{x}\)
\(=\frac{d y}{d x}=y\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x\right]\)
\(=\frac{d y}{d x}=(5 x)^{3 \cos 2 x}\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x\right]\)
\(\because \frac{d y}{d x}=(5 x)^{3 \cos 2 x}\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x\right]\)

Let \(y = \sin^{−1} (𝑥\sqrt{𝑥}), 0 \leq x \leq 1\)
Differentiating both sides with respect to \(x\), we get
Using chain rule we get
\(\frac{d y}{d x}=\frac{d}{d x} \sin ^{-1}(x \sqrt{x})\)
\(=\frac{d y}{d x}=\frac{1}{\sqrt{1-(x \sqrt{x})^{2}}} \times \frac{d}{d x}(x \sqrt{x})\)
\(=\frac{d y}{d x}=\frac{1}{\sqrt{1-x^{3}}} \times \frac{d}{d x}\left(x^{\frac{3}{2}}\right)=\frac{1}{\sqrt{1-x^{3}}} \times \frac{3}{2} x^{\frac{1}{2}}\)
\(=\frac{d y}{d x}=\frac{3 \sqrt{x}}{2 \sqrt{1-x^{3}}}\)
\(=\frac{d y}{d x}=\frac{3}{2} \sqrt{\frac{x}{\sqrt{1-x^{3}}}}\)
\(\therefore \frac{d y}{d x}=\frac{3}{2} \sqrt{\frac{x}{1-x^{3}}}\)

Let \( y=\frac{\cos ^{-1}}{\sqrt{2 x+7}},-2 < x < 2 \)
Differentiating both sides with respect to \( x \), we get
Using Quotient rule
\(\frac{d y}{d x}=\frac{\sqrt{2 x+7} \frac{d}{d x}\left(\cos ^{-1} \frac{x}{2}\right)-\left(\cos ^{-1} \frac{x}{2}\right) \frac{d}{d x}(\sqrt{2 x+7})}{(\sqrt{2 x+7})^{2}}\)
\(\frac{d y}{d x}=\frac{\sqrt{2 x+7}\left[\frac{-1}{\sqrt{1-\left(\frac{x}{2}\right)^{2}}} \times \frac{d}{d x}\left(\frac{x}{2}\right)\right]-\left(\cos ^{-1} \frac{x}{2}\right) \frac{d}{d x}(\sqrt{2 x+7})}{2 x+7}\)
\(\frac{d y}{d x}=\frac{\sqrt{2 x+7} \times-\frac{1}{\sqrt{4-x^{2}}}-\left(\cos ^{-1} \frac{x}{2}\right) \times \frac{2}{2 \sqrt{2 x+7}}}{2 x+7}\)
\(\frac{d y}{d x}=-\frac{\sqrt{2 x+7}}{\sqrt{4-x^{2} \times(2 x+7)}}-\frac{\cos ^{-1} \frac{x}{2}}{(\sqrt{2 x+7})(2 x+7)}\)
\(\therefore \frac{d y}{d x}=-\left[\frac{1}{\sqrt{4-x^{2}} \times \sqrt{2 x+7}}+\frac{\cos ^{-1} \frac{x}{2}}{(2 x+7)^{\frac{3}{2}}}\right]\)

Let \( y=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right), 0 < x < \frac{\pi}{2} \)
\(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\)
\(=\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})(\sqrt{1+\sin x}-\sqrt{1-\sin x})}\)
\(=\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{(1-\sin x)(1+\sin x)}}{(1+\sin x)(1-\sin x)}\)
\(=\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\frac{2+2 \sqrt{1-\sin^2 x}}{2 \sin x}\)
\(=\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\frac{1+\cos x}{\sin x}=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}=\cot \frac{x}{2}\)
Substituting the value of \( \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\cot \frac{x}{2} \) in \( y \).
\(\therefore y=\cot ^{-1}\left(\cot \frac{x}{2}\right)\)
\(\Rightarrow y=\frac{ x }{ 2 }\)
Differentiating both sides with respect to \(x \), we get
\(\frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}(x)\)
\(\therefore \frac{d y}{d x}=\frac{1}{2}\)

Let \( y=(\log x)^{\log x}, x > 1 \)
Taking logarithm on both sides
\(\Rightarrow \log y=\log (\log x) \log x=\log x \times \log (\log x)\)
Differentiating both sides with respect to \( x \), we get
\(\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}[\log x \times \log (\log x)]\)
\(\Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\log x) \times \frac{d}{d x}(\log x)+\log x \times \frac{d}{d x}[\log (\log x)]\)
\(\Rightarrow \frac{d y}{d x}=y\left[\log (\log x) \times \frac{1}{x}+\log x \times \frac{1}{\log x} \times \frac{d}{d x}(\log x)\right]\)
\(\Rightarrow \frac{d y}{d x}=y\left[\frac{1}{x} \log (\log x)+\frac{1}{x}\right]\)
\(\therefore \frac{d y}{d x}=(\log x)^{\log x}\left[\frac{1}{x}+\frac{\log (\log x)}{x}\right]\)

Let \( y=\cos (\mathrm{a} \cos x+\mathrm{b} \sin x) \)
\(a\) and \(b\) are some constants
\( y=\cos (a \cos x+b \sin x) \)
Differentiating both sides with respect to \(x \), we get
Using chain rule
\(\Rightarrow \frac{d y}{d x}=\frac{d}{d x} \cos (a \cos x+b \sin x)\)
\(\Rightarrow \frac{d y}{d x}=-\sin (a \cos +b \sin x) \times \frac{d}{d x}(a \cos x+b \sin x)\)
\(\Rightarrow \frac{d y}{d x}=-\sin (a \cos x+b \sin x) \times[a(-\sin x)+b \cos x]\)
\(\therefore \frac{d y}{d x}=(a \sin x-b \cos x) \times \sin (a \cos x+b \sin x)\)

Let \( y=\sin x-\cos x^{(\sin x-\cos x)}, \frac{\pi}{4} < x < \frac{3 \pi}{4} \)
Taking logarithm both sides, we get
\(\log y=\log [(\sin x-\cos x)(\sin x-\cos x)]\)
\(\Rightarrow \log y=(\sin x-\cos x) \times \log (\sin x-\cos x)\)
Differentiating both sides with respect to \( x \), we get
\(\frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}[(\sin x-\cos x) \times \log (\sin x-\cos x)]\)
\(\Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x) \times \frac{d}{d x}(\sin x-\cos )+(\sin x-\cos x) \times\)
\(\frac{d}{d x}(\sin x-\cos x)\)
\(\Rightarrow \frac{d y}{d x}=y[\log (\sin x-\cos x) \times(\cos x+\sin x)+(\sin x-\)
\(\left.\cos x) \times \frac{1}{(\sin x-\cos x)} \times \frac{d}{d x}(\sin x-\cos x)\right]\)
\(\Rightarrow \frac{d y}{d x}=y[\log (\sin x-\cos x) \times(\cos x+\sin x)+(\sin x-\cos x) \times\)
\(\left.\frac{1}{(\sin x-\cos x)} \times \frac{d}{d x}(\sin x-\cos x)\right]\)
\(\Rightarrow \frac{d y}{d x}=y[(\cos x+\sin x) \log (\sin x-\cos x)+(\cos x+\sin x)]\)
\(\therefore \frac{d y}{d x}=(\sin x-\cos x)(\sin x-\cos x)(\cos x+\sin x)[1+\log (\sin x-\cos x)]\)

Let \( y=x^{x}+x^{a}+a^{x}+a^{a} \), for some fixed \( a > 0 \) and \( x > 0 \)
And let \( x^{x}=u, x^{a}=v, a^{x}=w \) and \( a^{a}=s \)
Then \( y=\mathrm{u}+\mathrm{v}+\mathrm{w}+\mathrm{s} \)
\(\therefore \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}+\frac{d s}{d x} \ldots(i)\)
Now,
\(\mathrm{u}=x^{x}\)
Taking logarithm both sides, we get
\(\log \mathrm{u}=\log x^{x}\)
\(\Rightarrow \log u=x \log x\)
Differentiating both sides w.r.t. \(x\)
\(\Rightarrow \frac{1}{u} \frac{d y}{d x}=\log x \times \frac{d}{x}(x)+x \times \frac{d}{d x}(\log x)\)
\(\Rightarrow \frac{d y}{d x}=u\left[\log x+x \times \frac{1}{x}\right]\)
\(\Rightarrow \frac{d u}{d x}=x^{x}[\log x+1]=x^{x}(1+\log x) \ldots(ii)\)
\(\mathrm{v}=x^{\mathrm{a}}\)
Differentiating both sides with respect to \(x\)
\(\frac{d y}{d x}=\frac{d}{d x}\left(x^{a}\right)\)
\(\Rightarrow \frac{d v}{d x}=a x^{a-1}\ldots(iii)\)
\(\mathrm{w}=\mathrm{a}^{x}\)
Taking logarithm both sides
\(\log \mathrm{w}=\log \mathrm{a}x\)
\(\log \mathrm{w}=x \log \mathrm{a}\)
Differentiating both sides with respect to \(x\)
\(\frac{1}{w} \frac{d y}{d x}=\log a \times \frac{d}{d x}(x)\)
\(\Rightarrow \frac{d w}{d x}=w \log a\)
\(\Rightarrow \frac{d y}{d x}=a^{x} \log a \ldots(iv)\)
\(\mathrm{s}=\mathrm{a}^{\mathrm{a}}\)
Differentiating both sides with respect to \(x\)
\(\frac{d s}{d x}=0\ldots(v)\)
Putting (II), (III), (IV) and (V) in (I)
\(\frac{d y}{d x}=x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a+0\)
\(\therefore \frac{d y}{d x}=x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a\)

Let \( y=x^{x^{2}-3}+(x-3)^{x^{2}} \)
And let \( x^{x^{2}-3}=\mathrm{u} \&(x-3)^{x^{2}}=\mathrm{v} \)
\(\therefore y=\mathrm{u}+\mathrm{v}\)
Differentiating both sides w.r.t. \(x\) we get
\(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \ldots(i)\)
Now,
\(\mathrm{u}=x^{x^{2}-3}\)
Taking logarithm both sides
\(\log u=\log x^{x^{2} 3}\)
\(\Rightarrow \log u=(x 2-3) \log x\)
Differentiating w.r.t. \(x \), we get
\(\frac{1}{u} \frac{d y}{d x}=\log x \times \frac{d}{d x}\left(x^{2}-3\right)+\left(x^{2}-3\right) \times \frac{d}{d x}(\log x)\)
\(\Rightarrow \frac{d y}{d x}=u\left[\log x \times 2 x+\left(x^{2}-3\right) \times \frac{1}{x}\right]\)
\(\Rightarrow \frac{d u}{d x}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2 x \log x\right] \ldots (ii)\)
Also,
\(\mathrm{v}=(x-3)^{x^{2}}\)
Taking logarithm both sides
\(\log \mathrm{v}=\log (x-3)^{x^{2}}\)
\(\Rightarrow \log \mathrm{v}=x^{2} \log (x-3)\)
Differentiating both sides w.r.t. \(x\)
\(\frac{1}{v} \frac{d v}{d x}=\operatorname{loh}(x-3) \times \frac{d}{d x}\left(x^{2}\right) \times \frac{d}{d x}[\log (x-3)]\)
\(\Rightarrow \frac{d v}{d x}=v\left[\log (x-3) \times 2 x+x^{2} \times \frac{1}{(x-3)} \times \frac{d}{d x}(x-3)\right]\)
\(\Rightarrow \frac{d v}{d x}=(x-3)^{x^{2}}\left[2 x \log (x-3)+\frac{x^{2}}{(x-3)} \times 1\right]\)
\(\Rightarrow \frac{d v}{d x}=(x-3)^{x^{2}}\left[\frac{x^{2}}{(x-3)}+2 x \log (x-3)\right] \ldots \ldots(iii)\)
Substituting (II) and (III) in (I)
\(\therefore \frac{d y}{d x}=x^{x^{2}-3}\left[\frac{x^{2}-3}{x}+2 x \log x\right]+(x-3)^{x^{2}}\left[\frac{x^{2}}{(x-3)}+2 x \log (x-3)\right]\)

To find \( \frac{d y}{d x} \) we need to find out \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \)
So, \( \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \)
Given, \( y=12(1-\cos \mathrm{t}) \) and \( x=10(\mathrm{t}-\sin \mathrm{t}) \)
\(x=10(t-\sin t)\)
Differentiating with respect to \( t \).
\(\frac{d x}{d t}=\frac{d}{d t}[10(t-\sin t)]\)
\(\Rightarrow \frac{d x}{d t}=10 \times \frac{d y}{d x}(t-\sin t)=10(1-\cos t)\)
\(y=12(1-\cos \mathrm{t})\)
Differentiating with respect to \( t \).
\(\frac{d y}{d t}=\frac{d}{d x}[12(1-\cos t)]\)
\(\Rightarrow \frac{d y}{d x}=12 \times \frac{d y}{d x}(1-\cos t)=12 \times[0-(-\sin t)]=12 \sin t\)
\(\therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{12 \sin t}{10(1-\cos t)}=\frac{12 \times 2 \sin \frac{t}{2} \cos \frac{t}{2}}{10 \times 2 \sin ^{2} \frac{t}{2}}=\frac{6}{5} \cot \frac{t}{2}\)
\(\therefore \frac{d y}{d x}=\frac{6}{5} \cot \frac{t}{2}\)

Given,
\(y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}}\)
Differentiating with respect to \(x\)
\(\frac{d y}{d x}=\frac{d}{d x}\left[\sin^{-1} x+\sin^{-1} \sqrt{1-x^{2}}\right]\)
\(\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(\sin^{-1} x\right)+\frac{d}{d x}\left(\sin^{-1} \sqrt{1-x^{2}}\right)\)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-\left(\sqrt{1-x^{2}}\right)^{2}}} \times \frac{d}{d x}\left(\sqrt{1-x^{2}}\right)\)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-\left(1-x^{2}\right)}} \times \frac{d}{x}\left(\sqrt{1-x^{2}}\right)\)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{x} \times \frac{1}{2 \sqrt{1-x^{2}}} \times \frac{d}{d x}\left(1-x^{2}\right)\)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{2 x \sqrt{1-x^{2}}} \times(-2 x)\)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-x^{2}}}\)
\(\therefore \frac{d y}{d x}=0\)

Given, \( x \sqrt{1+y}+y \sqrt{1+x}=0 \)
\(x \sqrt{1+y}=-y \sqrt{1+x}=0\)
Now, squaring both sides, we get
\(\Rightarrow(x \sqrt{1+y})^{2}=(-y \sqrt{1+x})^{2}\)
\(\Rightarrow x^{2}(1+y)=y^{2}(1+x)\)
\(\Rightarrow x^{2}+x^{2} y=y^{2}+y^{2} x\)
\(\Rightarrow x^{2}-y^{2}=x^{2}-x^{2} y\)
\(\Rightarrow(x+y)(x-y)=x y(y-x)\)
\(\Rightarrow x+y=-\mathrm{xy}\)
\(\Rightarrow y+\mathrm{xy}=-x\)
\(\Rightarrow y(1+x)=-x\)
\(\Rightarrow y=-\frac{x}{(1+x)}\)
Differentiating both sides with respect to \( x \), we get
\(y=-\frac{x}{(1+x)}\)
Using Quotient Rule
\(y=-\frac{(1+x) \frac{d}{d x}(x)-x \frac{d}{d x}(1+x)}{\left(1+x^{2}\right)^{2}}=-\frac{(1+x)-x}{(1+x)^{2}}=-\frac{(1+x)-x}{(1+x)^{2}}=-\frac{1}{(1+x)^{2}}\)
\(\therefore \frac{d y}{d x}=-\frac{1}{(1+x)^{2}}\)
Hence, Proved

Given, \( (x-\mathrm{a})^{2}+(y-\mathrm{b})^{2}=\mathrm{c}^{2} \)
Differentiating with respect to \( x \), we get
\(\frac{d}{d x}\left[(x-a)^{2}\right]+\frac{d}{d x}\left[(y-b)^{2}\right]=\frac{d}{d x}\left(c^{2}\right)\)
\(\Rightarrow 2(x-\mathrm{a}) \times \frac{d}{d x}(x-a)+2(y-b) \times \frac{d}{d x}(y-b)=0\)
\(\Rightarrow 2(x-\mathrm{a}) \times 1+2(y-\mathrm{b}) \times \frac{d y}{d x}=0\)
\(\therefore \frac{d y}{d x}=-\frac{x-a}{y-b}\)
Differentiating again with respect to \(x\)
\(\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left[-\frac{(x-a)}{y-b}\right]\)
Using Quotient Rule
\(\Rightarrow \frac{d^{2} y}{d x^{2}}=-\left[\frac{(y-b) \times \frac{d}{d x}(x-a)-(x-a) \times \frac{d}{d x}(y-b)}{(y-b)^{2}}\right]\)
\(\Rightarrow \frac{d^{2} y}{d x^{2}}=-\left[\frac{(y-b)-(x-a) \times \frac{d}{d x}}{(y-b)^{2}}\right]\)
Substituting the value of \( \frac{d y}{d x} \) in the above equation
\(\Rightarrow \frac{d^{2} y}{d x^{2}}=-\left[\frac{(y-b)-(x-a) \times\left\{-\frac{(x-a)}{y-b}\right\}}{(y-b)^{2}}\right]\)
\(\Rightarrow \frac{d^{2} y}{d x^{2}}=-\left[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{3}}\right]\)
\(\therefore\left[\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}\right]^{\frac{3}{2}}=\frac{\left[1+\frac{(x-a)^{2}}{(y-b)^{2}}\right]^{\frac{3}{2}}}{-\left[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{3}}\right]}=\frac{\left[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{2}}\right]^{\frac{3}{2}}}{-\left[\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{3}}\right]}=\frac{\left[\frac{c^{3}}{(y-b)^{3}}\right]}{\frac{c^{2}}{(y-b)^{3}}}=-c\)
\(\therefore\left[\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}\right]^{\frac{3}{2}}=-\mathrm{c}, \text { which is independent of a and b }\)
Hence, Proved

Given, \( \cos y=x \cos (a+y) \)
Differentiating both sides with respect to \(x\)
\(\frac{d}{d y}[\cos y]=\frac{d}{d x}[x \cos (a+y)]\)
\(\Rightarrow-\sin y \frac{d y}{d x}=\cos (a+y) \times \frac{d}{d x}(x)+x \times \frac{d}{d x}[\cos (a+y)]\)
\(\Rightarrow-\sin y \frac{d y}{d x}=\cos (a+y)+x[-\sin (a+y)] \frac{d y}{d x}\)
\(\Rightarrow[x \sin (a+y)-\sin y] \frac{d y}{d x}=\cos (a+y) \ldots (i)\)
Since, \( \cos y=x \cos (a+y) \Rightarrow x=\cos y / \cos (a+y) \)
Substituting the value of \( x \) in (I)
\({\left[\frac{\cos y}{\cos (a+y)} \times \sin (a+y)-\sin y\right] \frac{d y}{d x}=\cos (a+y)}\)
\(\Rightarrow[\cos y \times \sin (a+y)-\sin y \times \cos (a+y)] \frac{d y}{d x}=\cos (a+y) \times \cos (a+y)\)
\(\Rightarrow \sin (a+y-y) \frac{d y}{d x}=\cos ^{2}(a+y)\)
\(\Rightarrow \sin a \times \frac{d y}{d x}=\cos ^{2}(a+y)\)
\(\Rightarrow \frac{d y}{d x}=\frac{\cos ^{2}(a+y)}{\sin a}\)
Hence, proved

Given, \( x=\mathrm{a}(\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t}) \) and \( y=(\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t}) \)
To find \( \frac{d y}{d x} \) we need to find out \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \)
So, \( \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \) and \( \frac{d^{2} y}{d x^{2}}=\frac{\frac{d y}{d x}}{d t} \times \frac{d t}{d x} \)
\(x=a(\cos t+t \sin t)\)
Differentiating with respect to \( t \).
\(\frac{d x}{d t}=\frac{d}{d t}[a(\cos t+t \sin t)]\)
\(\Rightarrow \frac{d x}{d t}=a \times \frac{d}{d t}(\cos t+t \sin t)=a\left[-\sin t+\sin t \times \frac{d}{d t}(t)+t \times\right.\)
\(\left.\frac{d}{d t}(\sin t)\right]\)
\(\Rightarrow \frac{d x}{d t}=a[-\sin t+\sin t+t \cos t]=a t \cos t\)
\(y=a(\sin t-t \cos t)\)
Differentiating with respect to \( t \).
\(\frac{d y}{d t}=a \frac{d}{d t}(\sin t-t \cos t)=a\left[\cos t-\left\{\cos t \times \frac{d}{d t}(t)+t \times \frac{d}{d t}(\cos t\}\right]\right.\)
\(\Rightarrow \frac{d y}{d t}=a[\cos t-\cos t+t \sin t]=a t \sin t\)
\(\therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a t \sin t}{a t \cos t}=\tan t\)
Differentiating \( \frac{ \mathrm{dy} }{ \mathrm{dx} } \) with respect to \(t\)
\(\frac{\frac{d y}{d x}}{d t}=\frac{d}{d t}(\tan t)=\sec ^{2} t\)
And \( \frac{d t}{d x}=\frac{1}{a t \cos t}=\frac{\sec t}{a t} \)
\(\frac{d^{2} y}{d x^{2}}=\frac{\frac{d y}{d x}}{d t} \times \frac{d t}{d x}\)
\(\Rightarrow \frac{d^{2} y}{d x^{2}}=\sec ^{2} t \times \frac{\sec t}{a t}\)
\(\therefore \frac{d^{2} y}{d x^{2}}=\frac{\sec ^{3} t}{a t}\)

\(|x|=\left\{\begin{array}{c}
x, \text { if } x \geq 0 \\
-x, \text { if } x < 0
\end{array}\right.\)
When, \( x \geq 0 \),
\(f(x)=|x|^{3}=x^{3}\)
So, \( \mathrm{f}^{\prime}(x)=3 x^{2} \)
And \( \mathrm{f}^{\prime}(x)=\mathrm{d}\left(\mathrm{f}^{\prime}(x)\right) / \mathrm{dx}=6 x \)
\(\therefore \mathrm{f}^{\prime}(x)=6 x\)
When \( x < 0 \),
\(\mathrm{f}(x)=|x|^{3}=(-x)^{3}=-x^{3}\)
\(\mathrm{f}^{\prime}(x)=-3 x^{2}\)
\(\mathrm{f}^{\prime}(x)=-6 x\)
\( \therefore \mathrm{f}^{\prime \prime}(x)=\left\{\begin{array}{c}
6 x, x \geq 0 \\
-6 x, x < 0
\end{array}\right.\)

To prove: \( \mathrm{P}(\mathrm{n}): \frac{d}{d x}\left(x^{n}\right)=\mathrm{nx}^{\mathrm{n}-1} \) for all positive integers n
For \( \mathrm{n}=1 \),
\(\mathrm{LHS}=\frac{d}{d x}(x)=1\)
\(\mathrm{RHS}=1 \times x 1-1=1\)
So, LHS \( = \) RHS
\( \therefore \mathrm{P}(1) \) is true.
\( \therefore \mathrm{P}(\mathrm{n}) \) is true for \( \mathrm{n}=1 \)
Let \( \mathrm{P}(\mathrm{k}) \) be true for some positive integer \( k \).
i.e. \( \mathrm{P}(\mathrm{k})=\frac{d y}{d x}\left(x^{k}\right)=k x^{k-1} \)
Now, to prove that \( P(k+1) \) is also true
\(\mathrm{RHS}=(\mathrm{k}+1) x(\mathrm{k}+1)-1\)
\(\mathrm{LHS}=\frac{d}{d x}\left(x^{k+1}\right)=\frac{d}{d x}\left(x \times x^{k}\right)\)
\(=x^{k} \times \frac{d}{d x}(x)+x \times \frac{d}{d x}\left(x^{k}\right)\)
\(=x^{k} \times 1+x \times k \times x^{k-1}\)
\(=x^{k}+k x^{k}\)
\(=(k+1) \times x^{k}\)
\(=(k+1) x^{(k+1)-1}\)
\(\therefore \text { LHS }=\text { RHS }\)
Thus, \( \mathrm{P}(\mathrm{k}+1) \) is true whenever \( \mathrm{P}(\mathrm{k}) \) is true.
Therefore, by the principle of mathematical induction, the statement \( \mathrm{P}(\mathrm{n}) \) is true for every positive integer \( n \).
Hence, proved.

\(\sin (A+B)=\sin A \cos B+\cos A \sin B\)
Differentiating with respect to \(x \), we get
\(\frac{d}{d x}[\sin (\mathrm{A}+\mathrm{B})]=\frac{d}{d x}(\sin \mathrm{A} \cos \mathrm{B})+\frac{d}{d x}(\cos \mathrm{A} \sin \mathrm{B})\)
\(\Rightarrow \cos (\mathrm{A}+\mathrm{B}) \frac{d}{d x}(\mathrm{~A}+\mathrm{B})=\cos \mathrm{B} \frac{d}{d x}(\sin \mathrm{A})+\sin \mathrm{A} \frac{d}{d x}(\cos \mathrm{B})+\sin\)
\(\mathrm{B} \frac{d}{d x}(\cos \mathrm{A})+\cos \mathrm{A} \frac{d}{d x}(\sin \mathrm{B})\)
\(\Rightarrow \cos (\mathrm{A}+\mathrm{B}) \frac{d}{d x}(\mathrm{~A}+\mathrm{B})=\cos \mathrm{B} \cos \mathrm{A} \frac{d A}{d x}+\sin \mathrm{A}(-\sin \mathrm{B}) \frac{d B}{d x}+\sin \mathrm{B}\)
\((-\sin \mathrm{A}) \frac{d A}{d x}+\cos \mathrm{A} \cos \mathrm{B} \frac{d B}{d x}\)
\(\Rightarrow \cos (\mathrm{A}+\mathrm{B}) \times\left[\frac{d A}{d x}+\frac{d B}{d x}\right]=(\cos \mathrm{A} \cos \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B}) \times\left[\frac{d A}{d x}+\frac{d B}{d x}\right]\)
\(\therefore \cos (\mathrm{A}+\mathrm{B})=\cos \mathrm{A} \mathrm{Cos} \mathrm{B}-\sin \mathrm{A} \sin \mathrm{B}\)

Considering the function
\(f(x)=|x|+|x+1|\)
The above function f is continuous everywhere, but is not differentiable at \( x=0 \) and \( x=-1 \)
\(\begin{array}{l}
\mathrm{f}(x)=\left\{\begin{array}{c}
-x-(x+1), x \leq-1 \\
-x+(x+1),-1 < x < 0 \\
x+(x+1), x \geq 0
\end{array}\right. \\
=\left\{\begin{array}{c}
-2 x-1, x \leq-1 \\
1,-1 < x < 0 \\
2 x+1, x \geq 0
\end{array}\right.
\end{array}\)
Now, checking continuity
CASE I: At \( x < -1 \)
\(f(x)=-2 x-1\)
\( \mathrm{f}(x) \) is a polynomial
\( \Rightarrow \mathrm{f}(x) \) is continuous [ \( \because \) Every polynomial function is continuous]
CASE II: \( x > 0 \)
\(f(x)=2 x+1\)
\( \mathrm{f}(x) \) is a polynomial
\( \Rightarrow \mathrm{f}(x) \) is continuous [ \( \because \) Every polynomial function is continuous]
CASE III: At \( -1 < x < 0 \)
\( \mathrm{f}(x)=1 \)
\( \mathrm{f}(x) \) is constant
\( \Rightarrow \mathrm{f}(x) \) is continuous
CASE IV: At \( x=-1 \)
\( \mathrm{f}(x)=\left\{\begin{array}{c}-2 x-1, x \leq-1 \\ 1,-1 < x < 0 \\ 2 x+1, x \geq 0\end{array}\right. \)
A function will be continuous at \( x=-1 \)
If \( \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(-1) \)
i.e. \( \lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{+}} f(x)=f(-1) \)
\( \mathrm{LHL}=\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{-}}-2 x-1 \)
Putting \( x=-1 \)
LHL \( =-2 \times(-1)-1=2-1=1 \)
\( \mathrm{RHL}=\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1^{+}} 1=1 \)
\( f(x)=-2 x-1 \)
\( \mathrm{f}(-1)=-2 \times(-1)-1=2-1=1 \)
so, \( \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(-1) \)
\( \Rightarrow \mathrm{f} \) is continuous.
CASE V: At \( x=0 \)
\(\mathrm{f}(x)=\left\{\begin{array}{c}
-2 x-1, x \leq-1 \\
1,-1 < x < 0 \\
2 x+1, x \geq 0
\end{array}\right.\)
A function will be continuous at \( x=0 \)
If \( \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(0) \)
i.e. \( \lim _{x \rightarrow 0^{-1}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0) \)
LHL \( =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} 1=1 \)
\( \mathrm{RHL}=\lim _{x \rightarrow 0^{+}}=\lim _{x \rightarrow 0^{+}} 2 x+1 \)
Putting \( x=0 \)
\( \mathrm{RHL}=2 \times 0+1=1 \)
\( \mathrm{f}(x)=2 x+1 \)
\( \mathrm{f}(0)=2 \times 0+1=0+1=1 \)
so, \( \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(0) \)
\( \Rightarrow \mathrm{f} \) is continuous.
Thus \( f(x)=|x|+|x+1| \) is continuous for all values of \( x \).
Checking differentiability
CASE I: At \( x < -1 \)
\(\begin{array}{l}
f(x)=-2 x-1 \\
f^{\prime}(x)=-2
\end{array}\)
\( \mathrm{f}(x) \) is polynomial.
\( \Rightarrow \mathrm{f}(x) \) is differentiable
CASE II: At \( x > 0 \)
\(f(x)=2 x+1\)
\(f^{\prime}(x)=2\)
\( \mathrm{f}(x) \) is polynomial.
\( \Rightarrow \mathrm{f}(x) \) is differentiable
CASE III: At \( -1 < x < 0 \)
\(f(x)=1\)
\( \mathrm{f}(x) \) is constant.
\( \Rightarrow \mathrm{f}(x) \) is differentiable
CASE IV: At \( x=-1 \)
\(\mathrm{f}(x)=\left\{\begin{array}{c}
-2 x-1, x \leq-1 \\
1,-1 < x < 0 \\
2 x+1, x \geq 0
\end{array}\right.\)
f is differentiable at \( x=-1 \) if
\(\mathrm{LHD}=\mathrm{RHD}=\mathrm{f}^{\prime}(-1)\)
i.e. \( \lim _{h \rightarrow-1^{-}} \frac{f(-1)-f(-1-h)}{h}=\lim _{h \rightarrow-1^{+}} \frac{f(-1+h)-f(-1)}{h}=f^{\prime}(-1) \)
\( \mathrm{LHD}=\lim _{h \rightarrow-1^{-}} \frac{f(-1)-f(-1-h)}{h}=\lim _{h \rightarrow-1^{-}} \frac{-2 \times(-1)-1-(-2 \times(-1-h)-1)}{h}= \) \( \lim _{h \rightarrow-1^{-}} \frac{2-1(2+2 h-1)}{h} \)
\( \mathrm{LHD}=\lim _{h \rightarrow-1^{-}} \frac{1-2 h-1}{h}=\lim _{h \rightarrow-1^{-}} \frac{-2 h}{h}=-2 \)
\( \mathrm{RHD}=\lim _{h \rightarrow-1^{+}} \frac{f(-1+h)-f(-1)}{h}=\lim _{h \rightarrow-1^{+}} \frac{1-(-2 \times(-1)-1}{h}=\lim _{h \rightarrow-1^{+}} \frac{1-1}{h}=0 \)
Since, LHD \( \neq \) RHD
\( \therefore \mathrm{f} \) is not differentiable at \( x=-1 \)
CASE V: At \( x=0 \)
\(\mathrm{f}(x)=\left\{\begin{array}{c}
-2 x-1, x \leq-1 \\
1,-1 < x < 0 \\
2 x+1, x \geq 0
\end{array}\right.\)
f is differentiable at \( x=0 \) if
\(\mathrm{LHD}=\mathrm{RHD}=\mathrm{f}^{\prime}(0)\)
i.e. \( \lim _{h \rightarrow 0^{-}} \frac{f(0)-f(0-h)}{h}=\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}=f^{\prime}(0) \)
\( \mathrm{LHD}=\lim _{h \rightarrow 0^{-}} \frac{f(0)-f(0-h)}{h}=\lim _{h \rightarrow 0^{-}} \frac{2 \times 0+1-1}{h}=0 \)
RHD \( =\lim _{h \rightarrow 0^{+}} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0^{+}} \frac{2 \times(0+h)+1-(2 \times 0+1)}{h}=\lim _{h \rightarrow-1^{+}} \frac{2 h+1-1}{h}=0 \)
Since, LHD \( \neq \) RHD
\( \therefore \mathrm{f} \) is not differentiable at \( x=0 \)
So, f is not differentiable at exactly two-point \( x=0 \) and \( x=1 \), but continuous at all points.

Let \( y=\left|\begin{array}{ccc}f(x) & g(x) & h(x) \\ 1 & m & n \\ a & b & c\end{array}\right| \)
Differentiation of determinant \( \mathrm{u}=\left|\begin{array}{ccc}e & f & g \\ h & i & j \\ k & l & m\end{array}\right| \) is given b
\(\begin{array}{l}
\frac{d y}{d x}=\left|\begin{array}{ccc}
\frac{d}{d x}(e) & \frac{d}{d x}(f) & \frac{d}{d x}(g) \\
h & i & j \\
k & l & m
\end{array}\right|+\left|\begin{array}{ccc}
e & f & g \\
\frac{d}{d x}(h) & \frac{d}{d x}(i) & \frac{d}{d x}(j) \\
k & l & m
\end{array}\right|+ \\
\left|\begin{array}{ccc}
e & f & g \\
h & i & j \\
\frac{d}{d x}(k) & \frac{d}{d x}(l) & \frac{d}{d x}(m)
\end{array}\right|+ \\
\frac{d y}{d x}=\left|\begin{array}{ccc}
\frac{d}{d x}(f(x)) & \frac{d}{d x}(g(x)) & \frac{d}{d x}(h(x)) \\
1 & m & n \\
a & b & b
\end{array}\right|+ \\
\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
\frac{d}{d x}(1) & \frac{d}{d x}(m) & \frac{d}{d x}(n) \\
a & b & c
\end{array}\right|+\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
1 & m & n \\
\frac{d}{d x}(a) & \frac{d}{d x}(b) & \frac{d}{d x}(c)
\end{array}\right| \\
\therefore \frac{d y}{d x}=\left|\begin{array}{cccc}
f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\
1 & m & n \\
a & b & c
\end{array}\right|+\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
0 & 0 & 0 \\
a & b & c
\end{array}\right|+ \\
\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
1 & m & n \\
0 & 0 & 0
\end{array}\right|
\end{array}\)
Since, \( \mathrm{a}, \mathrm{b}, \mathrm{c} \) and \( \mathrm{l}, \mathrm{m}, \mathrm{n} \) are constants so, their differentiation is zero.
Also in a determinant if all the elements of row or column turns to be zero then the value of determinant is zero.
\(\begin{array}{l}
\therefore\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
0 & 0 & 0 \\
a & b & c
\end{array}\right|=0 \text { and }\left|\begin{array}{ccc}
f(x) & g(x) & f(x) \\
1 & m & n \\
0 & 0 & 0
\end{array}\right| \\
\therefore \frac{d y}{d x}=\left|\begin{array}{ccc}
f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\
1 & m & n \\
a & b & c
\end{array}\right|
\end{array}\)
Hence, proved.

Given, \( y=e^{a \cos ^{-1} x} \)
Taking logarithm both sides, we get
\(\log y=\log e^{a \cos ^{-1} x}\)
\(\Rightarrow \log y=\mathrm{a} \cos -1 x \log \mathrm{e}\)
\(\Rightarrow \log y=\mathrm{a} \cos -1 x[\log \mathrm{e}=1]\)
Differentiating both sides with respect to \(x\)
\(\frac{1}{y} \frac{d y}{d x}=a \times-\frac{1}{\sqrt{1-x^{2}}}\)
\(\Rightarrow \frac{d y}{d x}=-\frac{a y}{\sqrt{1-x^{2}}}\)
Squaring both sides
\(\left(\frac{d y}{d x}\right)^{2}=\frac{a^{2} y^{2}}{1-x^{2}}\)
\(\Rightarrow(1-x^2)\left(\frac{d y}{d x}\right)^{2}=a^{2} y^{2}\)
Differentiating both sides
\(\Rightarrow\left(\frac{d y}{d x}\right)^{2} \frac{d y}{d x}\left(1-x^{2}\right)+\left(1-x^{2}\right) \times \frac{d}{d x}\left[\left(\frac{d y}{d x}\right)^{2}\right]=a^{2} \frac{d}{d x}\left(y^{2}\right)\)
\(\Rightarrow\left(\frac{d y}{d x}\right)^{2}(-2 x)+\left(1-x^{2}\right) \times 2 \times \frac{d}{d x} \times \frac{d^{2} y}{d x^{2}}=a^{2} \times 2 y \times \frac{d y}{d x}\)
\(\Rightarrow-x \times \frac{d y}{d x}+\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=a^{2} y\)
\(\therefore\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \times \frac{d y}{d x}-a^{2} y=0\)
Hence, proved