Differentiation (Part-38) | S N De

 

$~3.~$ Find the derivative w.r.t $~x~: $

$~(i)~x^{x^2}+a^{x^2}$

Sol. let $~~y=u+v \rightarrow(1)\\~[~\text{where}~u=x^{x^2},~v=a^{x^2}]$

Now, $~~~~u=x^{x^2} \\ \text{or,}~~\log u=x^2\log x \\ \text{or,}~~\frac 1u \frac{du}{dx}=2x\log x+x^2 \cdot \frac 1x \\ \text{or,}~~ \frac{du}{dx}=x^{x^2}[2x\log x+x]\rightarrow(2)$

Again, $~~v=a^{x^2} \\ \text{or,}~~\log v=x^2 \log a \\ \text{or,}~~ \frac 1v \cdot\frac{dv}{dx}=2x\log_e a \\ \text{or,}~~ \frac{dv}{dx}=2xv\log_ea \\ \text{or,}~~ \frac{dv}{dx}=2xa^{x^2}\log_ea\rightarrow(3)$

So, from $~(1),~(2),~(3)~$ we get,

$~~\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx} \\ \text{or,}~~ \frac{dy}{dx}=x^{x^2}\left(2x\log x+x\right)\\~~+2xa^{x^2}\log_ea~~\text{(ans.)}$

$\,(ii)~~\sin(\log x)+x^{\cos x}$

Sol. let $~~y=\sin(\log x)+x^{\cos x}=u+v,\text{(say)} \\ \therefore \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\rightarrow(1) $

Now, $~~~~u=\sin(\log x) \\ \text{or,}~~\frac{du}{dx}=\cos(\log x) \cdot \frac 1x \rightarrow(2) $

Again, $~~~v=x^{\cos x} \\ \text{or,}~~ \log v= \log (x^{\cos x}) \\ \text{or,}~~ \log v=\cos x \log x \\ \therefore \frac 1v \frac{dv}{dx}=-\sin x \log x+\cos x \cdot \frac 1x \\ \text{or,}~~ \frac{dv}{dx} =x^{\cos x }\left(-\sin x\log x+\frac{\cos x}{x}\right)\rightarrow(3) $

Hence, from $~(1),~(2),~(3)~$ we get,

$~\frac{dy}{dx}=\frac{\cos(\log x)}{x}\\~~~+x^{\cos x }\left(-\sin x\log x+\frac{\cos x}{x}\right)~~\text{(ans.)}$

To download full PDF containing full solution of DIFFERENTIATION (PART-1) [CLASS-XII] of Exercise-3A, click here.

$~(iii)~(\sin x)^{\cos x}+e^{3x}$

Sol. let  $~y=(\sin x)^{\cos x}+e^{3x}=u+v~\text{(say)} \\ \therefore \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx} \rightarrow(1)$

Now, $~u=(\sin x)^{\cos x} \\ \text{or,}~~\log u=\log (\sin x)^{\cos x} \\ \text{or,}~~ \log u=\cos x \log(\sin x) \\ \text{or,}~~ \frac 1u \frac{du}{dx}=-\sin x\log(\sin x)\\~~~~+ \cos x \cdot \frac{1}{\sin x} \cdot \cos x \\ \text{or,}~~ \frac{du}{dx}=(\sin x)^{\cos x}\\~~\times [-\sin x\log(\sin x)+\cos x \cdot \cot x] \rightarrow(2)$

Finally, $~~v=e^{3x} \Rightarrow \frac{dv}{dx}=3e^{3x} \rightarrow(3)$

$~\therefore~\frac{dy}{dx}=(\sin x)^{\cos x}[-\sin x\log(\sin x)\\+\cos x \cdot \cot x]+3e^{3x}~~\\~~~[\text{By (1),(2),(3)}]~~\text{(ans.)}$

$~(iv)~~x^x+(\sin x)^x$

Sol. let $~~y=x^x+(\sin x)^x=u+v~\text{(say)}\\ \therefore \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\rightarrow(1)$

Now, $~u=x^x \\ \text{or,}~~ \log u=\log(x^x) \\ \text{or,}~~ \log u=x\log x\\ \therefore \frac 1u \frac{du}{dx}=1.\log x+x\cdot \frac 1x \\ \text{or,}~~ \frac{du}{dx}=x^x(\log x+1)\rightarrow(2) $

Also, $~v=(\sin x)^x \\ \text{or,}~~ \log v=\log (\sin x)^x \\ \text{or,}~~ \log v=x\log(\sin x)\\ \therefore \frac 1v \frac{dv}{dx}=1 \cdot \log(\sin x)+x \cdot \frac{1}{\sin x}\cdot \cos x \\ \text{or,}~~ \frac{dv}{dx}=(\sin x)^x[\log(\sin x)+x \cot x]\rightarrow(3)$

Hence, from $~(1),~(2),~(3)~$ we get,

$~~\frac{dy}{dx}=x^x(\log x+1)\\~~~~+(\sin x)^x[\log(\sin x)+x \cot x]~~\text{(ans.)}$

$~(v)~~(\tan x)^{\cot x}+(\cot x)^{\tan x}$

Sol. let $~~y=(\tan x)^{\cot x}+(\cot x)^{\tan x}=u+v~\text{(say)}\\ \therefore \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\rightarrow(1)$

Now, $~u=(\tan x)^{\cot x} \\ \text{or,}~~ \log u=\log(\tan x)^{\cot x} \\ \text{or,}~~ \log u=\cot x\log \tan x\\ \therefore \frac 1u \frac{du}{dx}=-\csc^2x\log(\tan x)\\~~~+ \cot x .\frac{1}{\tan x}. \sec^2x \\ \therefore \frac 1u \frac{du}{dx}=-\csc^2x\log(\tan x)+\csc^2x\\~~~\left[\because ~\cot x \cdot \frac{1}{\tan x}\cdot \sec^2x\\=\frac{\sec^2x}{\tan^2x}=\frac{1}{\cos^2x \cdot \frac{\sin^2x}{\cos^2x}}=\csc^2x\right] \\ \text{or,}~~ \frac{du}{dx}=(\tan x)^{\cot x}\left[\csc^2x(1-\log(\tan x))\right]\rightarrow(2) $

Note : By $~\csc x~$ we mean $~\text{cosec}~x.$

Also, $~v=(\cot x)^{\tan x}\\ \text{or,}~~ \log v=\log (\cot x)^{\tan x} \\ \text{or,}~~ \log v=\tan x\log(\cot x)\\ \therefore \frac 1v \frac{dv}{dx}=\sec^2x \cdot \log(\cot x)\\~~~~+\tan x \cdot \frac{1}{\cot x}\cdot (-\csc^2x) \\ \therefore \frac 1v \frac{dv}{dx}=\sec^2x \cdot \log(\cot x)\\~~~+\tan^2x \cdot(-\csc^2x)\\ \text{or,}~~ \frac 1v\frac{dv}{dx} =\sec^2x \cdot \log(\cot x)-\sec^2x\\~~~\left[\because \tan^2x \cdot (-\csc^2x)\\=\frac{\sin^2x}{\cos^2x} \cdot \frac{-1}{\sin^2x}\\=-\sec^2x\right]\\ \text{or,}~~ \frac{dv}{dx}=(\cot x)^{\tan x}\left[\sec^2x(\log(\cot x)-1)\right]\rightarrow(3)$

Hence, from $~(1),~(2),~(3)~$ we get,

$~~\frac{dy}{dx}=(\tan x)^{\cot x}\left[\csc^2x(1-\log(\tan x))\right]\\~~~~+(\cot x)^{\tan x}\left[\sec^2x(\log(\cot x)-1)\right]~~\text{(ans.)}$