Ex 5.5 Class 12 Maths Ncert Solutions

Given: \( \cos x \cdot \cos 2 x \cdot \cos 3 x \)
Let \( y=\cos x \cdot \cos 2 x \cdot \cos 3 x \)
Taking \( \log \) on both sides, we get
\(\log y=\log (\cos x \cdot \cos 2 x \cdot \cos 3 x)\)
\(\Rightarrow \log y=\log (\cos x)+\log (\cos 2 x)+\log (\cos 3 x)\)
Now, differentiate both sides with respect to \(x\)
\(\frac{d}{d x}(\log y)=\frac{d}{d x} \log (\cos x)+\frac{d}{d x} \log (\cos 2 x)+\frac{d}{d x}(\log \cos 3 x)\)
\(=\frac{1}{y} \frac{d y}{d x}=\frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)+\frac{1}{\cos 2 x} \cdot \frac{d}{d x}(\cos 2 x)+\frac{1}{\cos 3 x} \frac{d}{d x}(\cos 3 x)\)
\(=\frac{d y}{d x}=y\left[-\frac{\sin x}{\cos x}-\frac{\sin 2 x}{\cos 2 x} \cdot \frac{d}{d x}(2 x)-\frac{\sin 3 x}{\cos 3 x} \frac{d}{d x}(3 x)\right]\)
\(=\frac{d y}{d x}=-\cos x \cdot \cos 2 x \cdot \cos 3 x[\tan x+\tan 2 x(2)+\tan 3 x(3)]\)
\(=\frac{d y}{d x}=-\cos x \cdot \cos 2 x \cdot \cos 3 x[\tan x+2 \tan 2 x+3 \tan 3 x]\)

Given: \( (\log x){ }^{\cos x} \)
Let \( x=(\log x)^{\cos x} \)
Taking \( \log \) on both sides, we get
\(\log y=\log (\log x)^{\cos x}\)
\(\Rightarrow \log y=\cos x \cdot \log (\log x)\)
Now, differentiate both sides with respect to \(x\)
\(\frac{d}{d x}(\log y)=\frac{d}{d x}[\cos x \cdot \log (\log x)]\)
\(=\frac{1}{y} \frac{d y}{d x}=\cos x \cdot \frac{d}{d x}(\log (\log x))+\log (\log x) \cdot \frac{d}{d x}(\cos x)\)
\(=\frac{d y}{d x}=y\left[\cos x \cdot \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)+\log (\log x) \cdot(-\sin x)\right]\)
\(=\frac{d y}{d x}=(\log x)^{\cos x}\left[\cos x \cdot \frac{1}{\log x} \cdot \frac{1}{x}+\log (\log x) \cdot(-\sin x)\right]\)
\(=\frac{d y}{d x}=(\log x)^{\cos x}\left[\frac{\cos x}{x \cdot \log x}-(\sin x) \cdot \log (\log x)\right]\)

Given: \( x^{x}-2^{\sin x} \)
Let \( y=x^{x}-2^{\sin x} \)
Let \( y=\mathrm{u}-\mathrm{v} \)
\(\Rightarrow \mathrm{u}=x^x \text { and } \mathrm{v}=2^{\sin } x\)
For, \( \mathrm{u}=x^{x} \)
Taking \( \log \) on both sides, we get
\(\log \mathrm{u}=\log x^{x}\)
\(\Rightarrow \log \mathrm{u}=x \cdot \log (x)\)
Now, differentiate both sides with respect to \(x \)
\(=\frac{d}{d x}(\log u)=\frac{d}{d x}[x \cdot \log (x)]\)
\(=\frac{1}{u} \frac{d u}{d x}=x \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}(x)\)
\(=\frac{d u}{d x}=u\left[x \cdot \frac{1}{x}+\log x \cdot(1)\right]\)
\(=\frac{d u}{d x}=x^{x}(1+\log x)\)
For, \( \mathrm{v}=2 \sin x \)
Taking \( \log \) on both sides, we get
\(\log \mathrm{v}=\log 2^{\sin x}\)
\(\Rightarrow \log \mathrm{v}=\sin x \cdot \log (2)\)
Now, differentiate both sides with respect to \(x\)
\(=\frac{d}{d x}(\log v)=\frac{d}{d x}[\sin x \cdot \log (2)]\)
\(=\frac{1}{v} \frac{d v}{d x}=\log 2 \cdot \frac{d}{d x}(\sin x)\)
\(=\frac{d v}{d x}=v[\log 2 \cdot(\cos x)]\)
\(=\frac{d v}{d x}=2^{\sin x} \cdot \cos x \log 2\)
Because, \( x=\mathrm{u}-\mathrm{v} \)
\(=\frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x}\)
\(\frac{ d y }{ d x }=x^{x}(1+\log x)-2 \sin x \cdot \cos x \cdot \log 2\)

Given: \( (x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4} \)
Let \( y=(x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4} \)
Taking \( \log \) on both sides, we get
\(\log y=\log \left((x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}\right)\)
\(\Rightarrow \log y=\log (x+3)^{2}+\log (x+4)^{3}+\log (x+5)^{4}\)
\(\Rightarrow \log y=2 \cdot \log (x+3)+3 \cdot \log (x+4)+4 \cdot \log (x+5)^{4}\)
Now, differentiate both sides with respect to \( x \)
\(=\frac{d}{d x}(\log y)=\frac{d}{d x}(2 \cdot \log (x+3))+\frac{d}{d x}(3 \cdot \log (x+4))+\frac{d}{d x}(4 \cdot \log (x+\)
5))
\(=\frac{1}{y} \frac{d y}{d x}=2 \cdot \frac{1}{x+3} \cdot \frac{d}{d x}(x+3)+3 \cdot \frac{1}{x+4} \cdot \frac{d}{d x}(x+4)+4 \cdot \frac{1}{x+5} \cdot \frac{d}{d x}(x+5)\)
\(=\frac{d y}{d x}=y\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]\)
\(=\frac{d y}{d x}=(x+3)^{2}(x+4)^{3}(x+5)^{4}\left[\frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)}\right]\)
\(=\frac{d y}{d x}=(x+3)^{1}(x+4)^{2}(x+5)^{3}\left[2\left(x^{2}+9 x+20\right)+3\left(x^{2}+8 x+\right.\right.\)
\(\left.15)+4\left(x^{2}+7 x+12\right)\right]\)
\(=(x+3)(x+4)^{2}(x+5)^{3}\left(9 x^{2}+70 x+133\right)\)

Given: \( \left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)} \)
Let \( x=\left(x+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)} \)
Also, Let \( x=\mathrm{u}+\mathrm{v} \)
\( =\mathrm{u}\left(x+\frac{1}{x}\right)^{x} \) and \( \mathrm{v}=x^{\left(1+\frac{1}{x}\right)} \)
for, \( \mathrm{u}=\left(x+\frac{1}{x}\right)^{x} \)
Taking \( \log \) on both sides, we get
\(\log \mathrm{u}=\log \left(x+\frac{1}{x}\right)^{x}\)
\(=\log \mathrm{u}=x \cdot \log \left(x+\frac{1}{x}\right)\)
Now, differentiate both sides with respect to \(x\)
\(\frac{d}{d x}(\log u)=\frac{d}{d x}\left[x \cdot \log \left(x+\frac{1}{x}\right)\right]\)
\(=\frac{1}{u}-\frac{d u}{d x}=x \cdot \frac{d}{d x}\left(\log \left(x+\frac{1}{x}\right)\right)+\log \left(x+\frac{1}{x}\right) \cdot \frac{d}{d x}(x)\)
\(=\frac{d u}{d x}=u\left[x \cdot \frac{1}{\left(x+\frac{1}{x}\right)} \cdot \frac{d}{d x}\left(x+\frac{1}{x}\right)+\log \left(x+\frac{1}{x}\right)\right]\)
\(=\frac{d u}{d x}=u\left[x \cdot \frac{1}{\left(x+\frac{1}{x}\right)} \cdot\left(\frac{d x}{d x}+\frac{d}{d x}\left(\frac{1}{x}\right)\right)+\log \left(x+\frac{1}{x}\right)\right]\)
\(=\frac{d u}{d x}=u\left[\frac{x}{\left(x+\frac{1}{x}\right)} \cdot\left(1-\frac{1}{x^{2}}\right)+\log \left(x+\frac{1}{x}\right)\right]\)
\(=\frac{d u}{d x}=u\left[\frac{x}{\left(x+\frac{1}{x}\right)} \cdot\left(\frac{x^{2}-1}{x^{2}}\right)+\log \left(x+\frac{1}{x}\right)\right]\)
\(=\frac{d u}{d x}=\left(x+\frac{1}{x}\right)^{x}\left[\left(\frac{x^{2}-1}{x^{2}+1}\right)+\log \left(x+\frac{1}{x}\right)\right]\)
for, \( \mathrm{v}=x^{\left(1+\frac{1}{x}\right)} \)
Taking \( \log \) on both sides, we get
\(\log \mathrm{v}=\log x^{\left(1+\frac{1}{x}\right)}\)
\(=\log \mathrm{v}=\left(1+\frac{1}{x}\right) \cdot \log x\)
Now, differentiate both sides with respect to \(x\)
\(=\frac{d}{d x}(\log v)=\frac{d}{d x}\left[\left(1+\frac{1}{x}\right) \cdot \log x\right]\)
\(=\frac{1}{v} \frac{d v}{d x}=\log x \cdot \frac{d}{d x}\left(1+\frac{1}{x}\right)+\left(1+\frac{1}{x}\right) \cdot \frac{d}{d x}(\log x)\)
\(=\frac{d v}{d x}=v\left[\log x \cdot\left(0-\frac{1}{x^{2}}\right)+\left(1+\frac{1}{x}\right) \cdot \frac{1}{x}\right]\)
\(=\frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left[-\frac{\log x}{x^{2}}+\left(\frac{1}{x}+\frac{1}{x^{2}}\right)\right]\)
\(=\frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left[\frac{-\log x+x+1}{x^{2}}\right]\)
\(=\frac{d v}{d x}=x^{\left(1+\frac{1}{x}\right)}\left[\frac{x+1-\log x}{x^{2}}\right]\)
Because, \( y=u+v \)
\(=\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
\(=\frac{d y}{d x}=\left(x+\frac{1}{x}\right)^{x}\left[\left(\frac{x^{2}-1}{x^{2}+1}\right)+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(1+\frac{1}{x}\right)}\left[\frac{x+1-\log x}{x^{2}}\right]\)

Given: \( (\log x)^{x}+x^{\log x} \)
Let \( y=(\log x)^x+x^{\log x} \)
Let \( y=\mathrm{u}+\mathrm{v} \)
\( \Rightarrow \mathrm{u}=(\log x)^{x} \) and \( \mathrm{v}=x^{\log x} \)
For, \( \mathrm{u}=(\log x)^{x} \)
Taking \( \log \) on both sides, we get
\(\log \mathrm{u}=\log (\log x)^{x}\)
\(\Rightarrow \log \mathrm{u}=x \cdot \log (\log (x))\)
Now, differentiate both sides with respect to \(x\)
\(\frac{d}{d x}(\log u)=\frac{d}{d x}[x \cdot \log (\log x)]\)
\(\left.=\frac{1}{u}-\frac{d u}{d x}=x \cdot \frac{d}{d x} \log (\log x)\right)+\log (\log x) \cdot \frac{d}{d x}(x)\)
\(=\frac{d u}{d x}=u\left[x \cdot \frac{1}{\log x} \frac{d}{d x}(\log x)+\log (\log x) \cdot(1)\right]\)
\(=\frac{d u}{d x}=(\log x)^{x}\left[\frac{x}{\log x} \cdot \frac{1}{x}+\log (\log x) \cdot(1)\right]\)
\(=\frac{d u}{d x}=(\log x)^{x}\left[\frac{1+\log (\log x) \cdot(\log x)}{\log x}\right]\)
\(=\frac{d u}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]\)
For, \( \mathrm{v}=\mathrm{x}^{\log x} \)
Taking \( \log \) on both sides, we get
\(\log \mathrm{v}=\log (x \log x)\)
\(\Rightarrow \log \mathrm{v}=\log x \cdot \log x\)
Now, differentiate both sides with respect to \(x\)
\(\frac{d}{d x}(\log v)=\frac{d}{d x}\left[(\log x)^{2}\right]\)
\(=\frac{1}{v} \frac{d v}{d x}=2 \cdot \log x \frac{d}{d x}(\log x)\)
\(=\frac{d v}{d x}=v\left[2 \cdot \frac{\log x}{x}\right]\)
\(=\frac{d v}{d x}=x^{\log x}\left[2 \cdot \frac{\log x}{x}\right]\)
\(=\frac{d v}{d x}=2 \cdot x^{\log x-1} \cdot \log x\)
Because, \( y=u+v \)
\(=\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
\(=\frac{d y}{d x}=(\log x)^{x-1}[1+\log x \cdot \log (\log x)]+2 \cdot x^{\log x-1} \cdot \log x\)

Given: \( (\sin x)^{x}+\sin ^{-1} \sqrt{x} \)
Let \( y=(\sin x)^{x}+ \sin ^{-1} \sqrt{x} \)
Let \( y=\mathrm{u}+\mathrm{v} \)
\(=\mathrm{u}=(\sin x)^{x} \text { and } \mathrm{v}=\sin ^{-1} \sqrt{x}\)
for, \( \mathrm{u}=(\sin x)^{x} \)
Taking \( \log \) on both sides, we get
\(\log \mathrm{u}=\log (\sin x)^{x}\)
Now, differentiate both sides with respect to \(x\)
\(=\frac{d}{d x}(\log u)=\frac{d}{d x}[x \cdot \log (\sin x)\)
\(\left.=\frac{1}{u} \frac{d u}{d x}=x \cdot \frac{d}{d x} \log (\sin x)\right)+\log (\sin x) \cdot \frac{d}{d x}(x)\)
\(=\frac{d u}{d x}=u\left[x \cdot \frac{1}{\sin x} \frac{d}{d x}(\sin x)+\log (\sin x) \cdot(1)\right]\)
\(=\frac{d y}{d x}=(\sin x)^{x}\left[\frac{x}{\sin x} \cdot \cos x+\log (\sin x) \cdot(1)\right]\)
\(=\frac{d y}{d x}=(\sin x)^{x}[x \cdot \cot x+\log \sin x]\)
for, \( \mathrm{v}= \sin ^{-1} \sqrt{x} \)
Now, differentiate both sides with respect to \(x\)
\(=\frac{d v}{d x}=\frac{d}{d x}\left[\sin^{-1} \sqrt{x}\right]\)
\(=\frac{d v}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{d}{d x}(\sqrt{x})\)
\(=\frac{d v}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2(\sqrt{x})}\)
\(=\frac{d v}{d x}= \frac{1}{2 \sqrt{x-x^{2}}} \quad \operatorname{Because}, y=\mathrm{u}+\mathrm{v}\)
\(=\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
\(=\frac{d y}{d x}=(\sin x)^{x}[x \cdot \cot x+\log \sin x]+\frac{1}{2 \sqrt{x-x^{2}}}\)

Given: \( x^{\sin x}+(\sin x)^{\cos x} \)
Let \( y=x^{\sin x}+(\sin x)^{\cos x} \)
Let \( y=u+v \)
\(\Rightarrow \mathrm{u}=x^{\sin x} \text { and } \mathrm{v}=(\sin x)^{\cos x}\)
For, \( \mathrm{u}=x^{\sin x} \)
Taking \( \log \) on both sides, we get
\(\log \mathrm{u}=\log (x \sin x)\)
\(\Rightarrow \log \mathrm{u}=\sin x \cdot \log (x)\)
Now, differentiate both sides with respect to \(x\)
\(=\frac{d}{d x}(\log u)=\frac{d}{d x}[\sin x \cdot \log x\)
\(=\frac{1}{u} \frac{d u}{d x}=\sin x \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}(\sin x)\)
\(=\frac{d u}{d x}=u\left[\sin x \cdot \frac{1}{x}+\log x \cdot \cos x\right]\)
\(=\frac{d u}{d x}=(x)^{\sin x}\left[\frac{\sin x}{x}+\log x \cdot \cos x\right]\)
For, \( \mathrm{v}=(\sin x)^{\cos x} \)
Taking \( \log \) on both sides, we get
\(\log \mathrm{v}=\log (\sin x)^{\cos x}\)
\(\Rightarrow \log \mathrm{v}=\cos x \cdot \log (\sin x)\)
Now, differentiate both sides with respect to \( x \)
\(\frac{d}{d x}(\log v)=\frac{d}{d x}[\cos x \cdot \log (\sin x)]\)
\(=\frac{1}{v} \frac{d v}{d x}=\cos x \cdot \frac{d}{d x} \log (\sin x)+\log \sin x \cdot \frac{d}{d x}(\cos x)\)
\(=\frac{d v}{d x}=v\left[\cos x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)+\log (\sin x) \cdot(-\sin x)\right]\)
\(=\frac{d v}{d x}=(\sin x)^{\cos x}\left[\frac{\cos x}{\sin x} \cdot \cos x+\log \sin x \cdot(-\sin x)\right]\)
\(=\frac{d y}{d x}=(\sin x)^{\cos x}[\cot x \cdot \cos x-\sin x \cdot \log \sin x] \text { Because, } x=\mathrm{u}+\mathrm{v}\)
\(=\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
\(=\frac{d y}{d x}=(x)^{\sin x}\left[\frac{\sin x}{x}+\log x \cdot \cos x\right]+(\sin x)^{\cos x}[\cot x \cdot \cos x-\sin x \cdot \log \sin x]\)

Given: \( x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1} \)
Let \( y=x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1} \)
Let \( y=u+v \)
\( =\mathrm{u}=x^{x \cos x} \) and \( \mathrm{v}=\frac{x^{2}+1}{x^{2}-1} \)
for, \( \mathrm{u}=x^{x \cos x} \)
Taking \( \log \) on both sides, we get
\(\log \mathrm{u}=\log x^{x \cos x}\)
\(\Rightarrow \log \mathrm{u}=x \cdot \cos x \cdot \log\)
Now, differentiate both sides with respect to \(x\)
\(\frac{d}{d x}(\log u)=\frac{d}{d x}[x \cdot \cos x \cdot \log x]\)
\(=\frac{1}{u} \frac{d u}{d x}=\cos x \log x \cdot \frac{d}{d x}(x)+x \cdot \log x \cdot \frac{d}{d x}(\cos x)+x \cdot \cos x \cdot \frac{d}{d x}(\log x)\)
\(=\frac{d u}{d x}=u\left[\cos x \cdot \log x+x \cdot \log x(-\sin x)+x \cdot \cos x \cdot\left(\frac{1}{x}\right)\right]\)
\(=\frac{d u}{d x}=x^{x \cos x}[\cos x \cdot \log x-x \cdot \log x \cdot \sin x+\cos x]\)
\(=\frac{d y}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \log x \cdot \sin x]\)
\(\text { for, } \mathrm{v}=\frac{x^{2}+1}{x^{2}-1}\)
Taking \( \log \) on both sides, we get
\(\log \mathrm{v}=\log \left(\frac{x^{2}+1}{x^{2}-1}\right)\)
\(\Rightarrow \log \mathrm{v}=\log \left(x^{2}+1\right)-\log \left(x^{2}-1\right)\)
Now, differentiate both sides with respect to \(x\)
\(\frac{d}{d x}(\log v)=\frac{d}{d x}\left[\log \left(x^{2}+1\right)-\log \left(x^{2}-1\right)\right]\)
\(=\frac{1}{v} \frac{d y}{d x}=\frac{1}{x^{2}+1} \cdot \frac{d}{d x}\left(x^{2}\right)-\frac{1}{x^{2}-1} \cdot \frac{d}{d x}\left(x^{2}\right)\)
\(=\frac{d y}{d x}=v \cdot\left[\frac{1}{x^{2}+1} \cdot(2 x)-\frac{1}{x^{2}-1} \cdot(2 x)\right]\)
\(=\frac{d y}{d x}=\left(\frac{x^{2}+1}{x^{2}-1}\right) \cdot\left[\frac{2 x\left(x^{2}-1\right)-2 x\left(x^{2}+1\right)}{\left(x^{2}+1\right)\left(x^{2}-1\right)}\right]\)
\(=\frac{d y}{d x}=\left(\frac{x^{2}+1}{x^{2}-1}\right) \cdot\left[\frac{-4 x}{\left(x^{2}+1\right)\left(x^{2}-1\right)}\right]\)
\(=\frac{d y}{d x}=\left[\frac{-4 x}{\left(x^{2}-1\right)^{2}}\right]\)
Because, \( y=u+v \)
\(=\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
\(=\frac{d y}{d x}=x^{x \cos x}[\cos x(1+\log x)-x \cdot \log x \cdot \sin x]-\left[\frac{4 x}{\left(x^{2}-1\right)^{2}}\right]\)

Given: \( (x \cos x)^{x}+(x \sin x)^{\frac{1}{x}} \)
Let \( y=(x \cos x)^{x}+(x \sin x)^{\frac{1}{x}} \)
Let \( y=u+v \)
\( =\mathrm{u}=(x \cos x)^{x} \) and \( \mathrm{v}=(x \sin x)^{\frac{1}{x}} \)
for, \( \mathrm{u}=(x \cos x)^{x} \)
Taking \( \log \) on both sides, we get
\(\log \mathrm{u}=\log (x \cos x)^{x}\)
\(\Rightarrow \log \mathrm{u}=x \cdot \log (x \cos x)\)
\(\Rightarrow \log \mathrm{u}=x(\log x+\log (\cos x))\)
\(\Rightarrow \log \mathrm{u}=x(\log x)+x(\log (\cos x))\)
Now, differentiate both sides with respect to \(x\)
\(=\frac{d y}{d x}(\log x)=\frac{d}{d x}[x \cdot \log (x)]+\frac{d}{d x}[x \cdot \log (\cos x)]\)
\(=\frac{1}{u} \frac{d u}{d x}=\left\{x \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}(x)\right\}+\left\{x \cdot \frac{d}{d x}(\log \cos x)+\right.\)
\(\left.\log \cos x \cdot \frac{d}{d x}(x)\right\}\)
\(=\frac{d u}{d x}=u\left[\left\{x \cdot \frac{1}{x}+\log x \cdot(1)\right\}+\left\{x \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)+\log \cos x \cdot(1)\right\}\right]\)
Taking \( \log \) on both sides, we get
\(\log \mathrm{v}=\log (x \sin x)^{\frac{1}{x}}\)
Now, differentiate both sides with respect to \(x\)
\(\frac{d}{d x}(\log v)=\frac{d}{d x}\left[\frac{1}{x} \cdot(\log x)\right]+\frac{d}{d x}\left[\frac{1}{x} \cdot \log (\sin x)\right]\)
\(=\frac{1}{v} \frac{d y}{d x}=\left\{\frac{1}{x} \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}\left(\frac{1}{x}\right)\right\}+\left\{\frac{1}{x} \cdot \frac{d}{d x}(\log \sin x)+\right.\)
\(\left.\log \sin x \cdot \frac{d}{d x}\left(\frac{1}{x}\right)\right\}\)
\(=\frac{d y}{d x}=v\left[\left\{\frac{1}{x} \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}\left(\frac{1}{x}\right)\right\}+\left\{\frac{1}{x} \cdot \frac{d}{d x}(\log \sin x)+\right.\right.\)
\(\left.\left.\log \sin x \cdot \frac{d}{d x}\left(\frac{1}{x}\right)\right\}\right]\)
\(=\frac{d y}{d x}=(x \sin x)^{\frac{1}{x}}\left[\left\{\frac{1}{x^{2}}(1-\log x)\right\}+\left\{\frac{\cos x}{x \cdot \sin x} \cdot-\frac{\log \sin x}{x^{2}}\right\}\right]\)
\(=\frac{d y}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log x}{x^{2}}+\frac{\cot x}{x}-\frac{\log \sin x}{x^{2}}\right]\)
\(=\frac{d y}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1-\log x+x \cot x-\log \sin x}{x^{2}}\right]\)
\(=\frac{d y}{d x}=(x \sin x)^{\frac{1}{x}}\left[\frac{1+x \cot x-\log (x \cdot \sin x)}{x^{2}}\right]\)
\(=\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\)
\(=\frac{d y}{d x}=(x \cos x)^{x}[1-x \cdot \tan x+\log (x \cdot \cos x)]+ (x \sin x)^{\frac{1}{x}}\left[\frac{1+x \cot x-\log (x \cdot \sin x)}{x^{2}}\right]\)

Given: \( x^{y}+y^{x}=1 \)
Let \( y=x^{y}+y^{x}=1 \)
Let \( \mathrm{u}=x^{y} \) and \( \mathrm{v}=y^{x} \)
Then, \( \Rightarrow \mathrm{u}+\mathrm{v}=1 \)
\(=\frac{d u}{d x}+\frac{d v}{d x}=0\)
For, \( \mathrm{u}=x y \)
Taking \( \log \) on both sides, we get
\( \log \mathrm{u}=\log x y \)
\(\Rightarrow \log \mathrm{u}=x \cdot \log (x)\)
Now, differentiate both sides with respect to \(x\)
\(=\frac{d}{d x}(\log u)=\frac{d}{d x}[y \cdot \log (x)]\)
\(=\frac{1}{u} \frac{d u}{d x}=\left\{y \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}(y)\right\}\)
\(=\frac{d u}{d x}=u\left[y \cdot \frac{1}{x}+\log x \cdot\left(\frac{d y}{d x}\right)\right]\)
\(=\frac{d y}{d x}=x^{y}\left[\frac{y}{x}+\log x \cdot\left(\frac{d y}{d x}\right)\right]\)
For, \( \mathrm{v}=y^{x} \)
Taking \( \log \) on both sides, we get
\(\log \mathrm{v}=\log y^{x}\)
\(\Rightarrow \log \mathrm{v}=x \cdot \log (y)\)
Now, differentiate both sides with respect to \(x\)
\(=\frac{d}{d x}(\log v)=\frac{d}{d x}[x \cdot \log (y)]\)
\(=\frac{1}{v} \frac{d v}{d x}=\left\{x \cdot \frac{d}{d x}(\log y)+\log y \cdot \frac{d}{d x} x\right\}\)
\(=\frac{d v}{d x}=v\left[x \cdot \frac{1}{y} \cdot \frac{d y}{d x}+\log y \cdot\left(\frac{d y}{d x}\right)\right]\)
\(=\frac{d y}{d x}=y^{x}\left[\frac{x}{y} \cdot \frac{d y}{d x}+\log y\right]\)
because, \( \frac{d u}{d x}+\frac{d v}{d x}=0 \)
\(\text { so, } x^{y}\left[\frac{y}{x}+\log x \cdot\left(\frac{d y}{d x}\right)\right]+y^{x}\left[\frac{x}{y} \cdot \frac{d y}{d x}+\log y\right]=0\)
\(=\left(x^{y} \log x+x y^{x-1}\right) \cdot \frac{d y}{d x}+\left(y x^{y-1}+y^{x} \log y\right)=0\)
\(=\left(x^{y} \log x+x y^{x-1}\right) \cdot \frac{d y}{d x}=-\left(y x^{y-1}+y^{x} \log y\right)\)
\(=\frac{d y}{d x}=-\frac{\left(y x^{y-1}+y^{x} \log y\right)}{\left(x^{y} \log x+x y^{x-1}\right)}\)

Given: \( y^{x}=x^{y} \)
Taking \( \log \) on both sides, we get
\(\log yx=\log x^{y}\)
\(\Rightarrow x \log y=x \log x\)
Now, differentiate both sides with respect to \(x\)
\(x \cdot \frac{d}{d x} \log y+\log y \cdot \frac{d}{d x} x=y \cdot \frac{d}{d x} \log x+\log x \cdot \frac{d}{d x} y\)
\(x \cdot \frac{1}{y} \cdot \frac{d y}{d x}+\log y \cdot(1)=y \cdot \frac{1}{x}+\log x \cdot \frac{d y}{d x}\)
\(\frac{x}{y} \cdot \frac{d y}{d x}-\log x \cdot \frac{d y}{d x}=y \cdot \frac{1}{x}-\log y\)
\(=\frac{d y}{d x}\left(\frac{x}{y}-\log x\right)=\frac{y-x \log y}{x}\)
\(=\frac{d y}{d x}\left(\frac{x-y \log x}{y}\right)=\frac{y-x \log y}{x}\)
\(=\frac{d y}{d x}=\frac{y}{x}\left(\frac{y-x \log y}{x-y \log x}\right)\)

Given: \( (\cos x)^{y}=(\cos x)^x \)
Taking \( \log \) on both sides, we get
\(\log (\cos x)^{y}=\log (\cos y)^{x}\)
\(\Rightarrow y \log (\cos x)=x \log (\cos y)\)
Now, differentiate both sides with respect to \(x \)
\(y\cdot \frac{d}{d x} \log (\cos x)+\log (\cos x) \cdot \frac{d}{d x} y=x \cdot \frac{d}{d x} \log (\cos y)+\log \cos y \cdot \frac{d}{d x} x\)
\(=y \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)+\log (\cos x) \cdot \frac{d y}{d x}=x \cdot \frac{1}{\cos y} \cdot \frac{d}{d x}(\cos y)+\)
\(\log (\cos y) \cdot \frac{d y}{d x}\)
\(=\frac{y}{\cos x} \cdot(-\sin x)+\log (\cos x) \cdot \frac{d y}{d x}=\frac{x}{\cos y} \cdot(-\sin y) \cdot \frac{d y}{d x}+\)
\(\log (\cos y) \cdot(1)\)
\(=\frac{d y}{d x}\left(\frac{x \cdot \sin y}{\cos y}+\log (\cos x)\right)=y \cdot \frac{\sin x}{\cos x}+\log (\cos y)\)
\(=\frac{d y}{d x}(x \tan x+\log (\cos x))=y \cdot \tan x+\log (\cos y)\)
\(=\frac{d y}{d x}=\left(\frac{y \cdot \tan x+\log (\cos y)}{x \cdot \tan x+\log (\cos x)}\right)\)

Given: \( xy=\mathrm{e}^{(x-y)} \)
Taking \( \log \) on both sides, we get
\(\log (x y)=\log (e(x-y))\)
\(\Rightarrow \log x+\log y=(x-y) \log e\)
\(\Rightarrow \log x+\log y=(x-y) .1\)
\(\Rightarrow \log x+\log y=(x-y)\)
Now, differentiate both sides with respect to \(x\)
\(\frac{d}{d x} \log x+\frac{d}{d x} \log y=\frac{d}{d x} x-\frac{d}{d x} y\)
\(\frac{1}{x}+\frac{1}{y} \frac{d y}{d x}=1-\frac{d y}{d x}\)
\(\left(1+\frac{1}{y}\right) \frac{d y}{d x}=1-\frac{1}{x}\)
\(\frac{1+y}{y} \frac{d y}{d x}=\frac{x-1}{x}\)
\(\frac{d y}{d x}=\frac{y(x-1)}{x(1+y)}\)

Given: \( \mathrm{f}(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right) \)
Taking \( \log \) on both sides, we get
\(\log \mathrm{f}(x)=\log (1+x)+\log \left(1+x^{2}\right)+\log \left(1+x^{4}\right)+\log \left(1+x^{8}\right)\)
Now, differentiate both sides with respect to \(x\)
\(\frac{d}{d x} \log f(x)=\frac{d}{d x} \log (1+x)+\frac{d}{d x} \log \left(1+x^{2}\right)+\frac{d}{d x} \log \left(1+x^{4}\right)+ \frac{d}{d x} \log \left(1+x^{8}\right)\)
\(=\frac{1}{f(x)} \cdot \frac{d}{d x}[f(x)]\)
\(=\frac{1}{1+x} \cdot \frac{d}{d x}(1+x)+\frac{1}{1+x^{2}} \frac{d}{d x}\left(1+x^{2}\right)+\frac{1}{1+x^{4}} \frac{d}{d x}\left(1+x^{4}\right)+\frac{1}{1+x^{8}} \frac{d}{d x}(1+x^{8})\)
\(=\mathrm{f}^{\prime}(x)=\mathrm{f}(x)=\left[\frac{1}{1+x}+\frac{1}{1+x^{2}} \cdot(2 x)+\frac{1}{1+x^{4}} \cdot\left(4 x^{3}\right)+\frac{1}{1+x^{8}}\left(8 x^{7}\right)\right]\)
\(=f^{\prime}(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)\left[\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{4 x^{3}}{1+x^{4}}+ \frac{8 x^{7}}{1+x^{8}}\right]\)
\(=f^{\prime}(x)=(1+1)\left(1+1^{2}\right)\left(1+1^{4}\right)\left(1+1^{8}\right)\left[\frac{1}{1+1}+\frac{2(1)}{1+1}+\frac{4(1)^{3}}{1+(1)^{4}}+ \frac{8(1)^{7}}{1+(1)^{8}}\right]\)
\(=f^{\prime}(1)=(2)(2)(2)(2)\left[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right]\)
\(=f^{\prime}(1)=16\left[\frac{1+2+4+8}{2}\right]\)
\(=f^{\prime}(1)=16\left(\frac{15}{2}\right)\)
\(=f^{\prime}(1)=120\)

Given: \( \left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right) \)
Let \( y=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right) \)
(i) By applying product rule differentiate both sides with respect to \(x\)
\(\frac{d y}{d x}=\frac{d y}{d x}\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)\)
\(=\frac{d y}{d x}=\left(x^{3}+7 x+9\right) \cdot \frac{d}{d x}\left(x^{2}-5 x+8\right)+\left(x^{2}-5 x+8\right) \cdot \frac{d}{d x}\left(x^{3}+7 x+9\right)\)
\(=\frac{d y}{d x}=\left(x^{3}+7 x+9\right) \cdot(2 x-5)+\left(x^{2}-5 x+8\right) \cdot\left(3 x^{2}+7\right)\)
\(=\frac{d y}{d x}=2 x^{4}+14 x^{2}+18 x-5 x^{3}-35 x-45+3 x^{4}+7 x^{2}-15 x^{3}- 35 x+24 x^{2}+56\)
\(=\frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11 \ldots(1)\)
(ii) by expanding the product to obtain a single polynomial
\(y=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)\)
\(y=x^{5}+7 x^{3}+9 x^{2}-5 x^{4}-35 x^{2}-45 x+8 x^{3}+56 x+72\)
\(y=x^{5}-5 x^{4}+15 x^{3}-26 x^{2}+11 x+72\)
Now, differentiate both sides with respect to \(x\)
\(\frac{d y}{d x}=\frac{d}{d x}\left(x^{5}\right)-\frac{d}{d x}\left(5 x^{4}\right)+\frac{d}{d x}\left(15 x^{3}\right)-\frac{d}{d x}\left(26 x^{2}\right)+\frac{d}{d x}(11 x)+\)
\(\frac{d}{d x}(72)\)
\(\frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11 \ldots(2)\)
(iii) by logarithmic differentiation
\(y=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)\)
Taking \( \log \) on both sides, we get
\(\log y=\log \left(\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)\right)\)
\(\log y=\log \left(x^{2}-5 x+8\right)+\log \left(x^{3}+7 x+9\right)\)
Now, differentiate both sides with respect to \(x\)
\(\frac{d y}{d x}(\log y)=\frac{d}{d x} \log \left(x^{2}-5 x+8\right)+\frac{d}{d x} \log \left(x^{3}+7 x+9\right)\)
\(=\frac{1}{y} \frac{d}{d x}(y)=\left[\frac{1}{\left(x^{2}-5 x+8\right)} \cdot \frac{d}{d x}\left(x^{2}-5 x+8\right)+\frac{1}{\left(x^{3}+7 x+9\right)} \cdot \frac{d}{d x}\left(x^{3}+7 x+9\right) \right]\)
\(=\frac{1}{y} \frac{d}{d x}(y)=\left[\frac{1}{\left(x^{2}-5 x+8\right)} \cdot(2 x-5)+\frac{1}{\left(x^{3}+7 x+9\right)} \cdot\left(3 x^{2}+7\right)\right]\)
\(=\frac{d}{d x}(y)=y \cdot\left[\frac{(2 x-5)}{\left(x^{2}-5 x+8\right)}+\frac{\left(3 x^{2}+7\right)}{\left(x^{3}+7 x+9\right)}\right]\)
\(=\frac{d}{d x}(y)=y \cdot\left[\frac{(2 x-5)\left(x^{3}+7 x+9\right)+\left(3 x^{2}+7\right)\left(x^{2}-5 x+9\right)}{\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)}\right]\)
\(=\frac{d}{d x}(y)=y \cdot\left[\frac{2 x^{4}+14 x^{2}+18 x-5 x^{3}-35 x-45+3 x^{4}-15 x^{3}+24 x^{2}+7 x^{2}-35 x+56}{(x-5 x+8)(x+7 x+9)}\right]\)
\(=\frac{d}{d x}(y)=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right) \cdot\left[\frac{5 x^{4}-20 x^{3}-45 x^{2}-52 x+11}{\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)}\right]\)
\(=\frac{d y}{d x}=5 x^{4}-20 x^{3}+45 x^{2}-52 x+11 \ldots . .(3)\)
From equation (i), (ii) and (iii), we can say that value of given function after differentiating by all the three methods is same.

To prove: \( \frac{d}{d x}(u \cdot v \cdot w)=\frac{d u}{d x} v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot \frac{d w}{d x} \)
Let \( y=\mathrm{u} . \mathrm{v} \). w=u. (v. w)
(a) by applying product rule differentiate both sides with respect to \(x\)
\(\frac{d y}{d x}=(v \cdot w) \cdot \frac{d u}{d x}+u \cdot \frac{d}{d x}(v \cdot w)\)
\(=\frac{d y}{d x}=(v \cdot w) \cdot \frac{d u}{d x}+u \cdot\left[v \cdot \frac{d}{d x}(w)+w \cdot \frac{d}{d x}(v)\right]\)
\(=\frac{d y}{d x}=(v \cdot w) \cdot \frac{d u}{d x}+(u \cdot v) \cdot \frac{d w}{d x}+(u \cdot w \cdot) \cdot \frac{d v}{d x}\)
(b) Taking log on both sides, we get
as, \( y=\mathrm{u} . \mathrm{v} . \mathrm{w} \)
\(\log y=\log (u . v . w)\)
\(\log y=\log u+\log v+\log w\)
Now, differentiate both sides with respect to \(x \)
\(=\frac{d}{d x}(\log y)=\frac{d}{d x} \log u+\frac{d}{d x} \log v+\frac{d}{d x} \log w\)
\(=\frac{1}{y} \cdot \frac{d}{d x}(y)=\frac{1}{u} \cdot \frac{d}{d x}(u)+\frac{1}{v} \cdot \frac{d}{d x}(v)+\frac{1}{w} \cdot \frac{d}{d x}(w)\)
\(=\frac{d y}{d x}(y)=y\left[\frac{1}{u} \cdot \frac{d u}{d x}+\frac{1}{v} \cdot \frac{d v}{d x}+\frac{1}{w} \cdot \frac{d w}{d x}\right]\)
\(=\frac{d y}{d x}=u \cdot v \cdot w\left[\frac{1}{u} \cdot \frac{d u}{d x}+\frac{1}{v} \frac{d v}{d x}+\frac{1}{w} \cdot \frac{d w}{d x}\right]\)
\(=\frac{d y}{d x}=v \cdot w \cdot \frac{d u}{d x}+u \cdot w \cdot \frac{d v}{d x}+u \cdot v \cdot \frac{d w}{d x}\)
From equation (i), (ii) and (iii), we can say that value of given function after differentiating by all the three methods is same.