Differentiation (Part-38) | S N De

 

$~2(i)~\log_x\sin x$

Sol. let $~~~y=\log_x\sin x=\frac{\log_e\sin x}{\log_e x} \\ \therefore \frac{dy}{dx}\\=\frac{(\log_e x) \frac{d}{dx}(\log_e \sin x)-\log_e \sin x \frac{d}{dx}(\log_e x)}{(\log_e x)^2}\\=\frac{(\log_e x) \frac{1}{\sin x}\cdot \cos x-\frac{\log_e \sin x}{x}}{(\log_e x)^2}\\=\frac{( \log_e x) ~x \cot x-\log_e \sin x}{x(\log_e x)^2}~~\text{(ans.)}$

$~(ii)~~\log_{\sin x}x$

Sol.  let $~~y=\log_{\sin x}x=\frac{\log_e x}{\log_e \sin x} \\ \therefore \frac{dy}{dx}\\=\frac{(p) \frac{d}{dx}(\log_e x)-(\log_e x)\frac{d}{dx}(p)}{(p)^2}\\=\frac{(\log_e \sin x)\cdot \frac 1x-(\log_e x) \cdot \frac{1}{\sin x}\cdot \cos x}{(\log_e \sin x)^2}\\=\frac{\log_e \sin x-x(\log_e x)\cot x}{x(\log_e \sin x)^2}~~\text{(ans.)}$

$~(iii)~~\log_x \tan x$

Sol. let $~y=\log_x \tan x=\frac{\log_e \tan x}{\log_e x}\\ \therefore~\frac{dy}{dx}\\=\frac{(\log_e x)\frac{d}{dx}(\log_e\tan x)-(\log_e \tan x)\frac{d}{dx}(\log_e x)}{(\log_e x)^2}\\=\frac{(\log_e x)\cdot \frac{1}{\tan x} \cdot \sec^2x-(\log_e\tan x)\cdot \frac 1x }{(\log_e x)^2}\\=\frac{x(\log_e x)\cdot \frac{\cos x\sec^2x}{\sin x}-\log_e \tan x }{x(\log_e x)^2}\\=\frac{x(\log_e x)\frac{2\sec x}{2\sin x}-\log_e(\tan x)}{x(\log_e x)^2}\\=\frac{2x \csc x(\log_e x)-\log_e(\tan x) }{x(\log_e x)^2}~[*]~\text{(ans.)}$

Note [*]: $~~\frac{2\sec x}{2\sin x}\\=\frac{2}{2\sin x\cos x}\\=\frac{2}{\sin2x}\\=2\csc x\\~[~\text{here by }~~\csc x ~~\text{we mean  cosec }~x.~]$

$~(iv)~~\log_{\sin x} (\sec x)+10^{x^2}$

Sol. let $~~y=\log_{\sin x} (\sec x)+10^{x^2}\\ \text{or,}~~ y=u+v~~\text{(say)} \\ \therefore \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\rightarrow(1)$

$~~u=\log_{\sin x} (\sec x)=\frac{\log_e(\sec x)}{\log_e \sin x} \\ \therefore \frac{du}{dx}\\=\frac{(\log_e \sin x) \frac{d}{dx} (\log_e \sec x)-(\log_e \sec x)\frac{d}{dx}(\log_e \sin x)}{(\log_e \sin x)^2}\\=\frac{(\log_e \sin x) \frac{1}{\sec x}\cdot \sec x\tan x-(\log_e \sec x) \frac{1}{\sin x}\cdot \cos x}{(\log_e \sin x)^2}\\=\frac{\tan x(\log_e \sin x)+\cot x\cdot \log_e(\sec x)^{-1}}{(\log_e \sin x)^2}\\=\frac{\tan x  (\log_e \sin x)+\cot x\cdot \log_e (\cos x)}{(\log_e \sin x)^2}\rightarrow(2)$

$~~v=10^{x^2}\\ \text{or,}~~ \log v= \log(10^{x^2}) =x^2 \log_e 10\\ \therefore \frac 1v \frac{dv}{dx}= 2x \log_e 10 \\ \therefore \frac{dv}{dx}=10^{x^2}(2x \log_e 10) \rightarrow(3)$

Hence from $\,(1),\,(2),\,(3)\,$ we get, the required derivative 

$~~=\frac{\tan x  (\log_e \sin x)+\cot x\cdot \log_e (\cos x)}{(\log_e \sin x)^2}\\~~~~~~~+10^{x^2}(2x \log_e 10)~~\text{(ans.)}$

$~3.~~\tan^{-1}\left(\frac{\cos x}{1+\sin x}\right)+\sin(\log x)$

Sol. let $~y=\tan^{-1}\left(\frac{\cos x}{1+\sin x}\right)+\sin(\log x)\rightarrow(1)$

Now, $~~~~~~~~\frac{\cos x}{1+\sin x}\\=\frac{\cos\left(2 \cdot \frac x2\right)}{\cos^{2}(x/2)+\sin^2(x/2)+2\sin (x/2)\cos(x/2)}\\=\frac{\cos^2(x/2)-\sin^2(x/2)}{(\cos(x/2)+\sin(x/2))^2}\\=\frac{[\cos(x/2)+\sin(x/2)][\cos(x/2)-\sin(x/2)]}{[\cos(x/2)+\sin(x/2)]^2}\\=\frac{\cos(x/2)-\sin(x/2)}{\cos(x/2)-\sin(x/2)}\\=\frac{1-\tan(x/2)}{1+\tan(x/2)}\\=\frac{\tan(\pi/4)-\tan(x/2)}{1+\tan(\pi/4)\tan(x/2)}\\=\tan\left(\frac{\pi}{4}-\frac x2\right)\rightarrow(2)$

Hence from $\,(1),\,(2)\,$  we get,

$~~y=\tan^{-1} \tan\left(\frac{\pi}{4}-\frac x2\right)+\sin(\log x) \\ \text{or,}~~y=\frac{\pi}{4}-\frac x2+\sin(\log x) \\ \therefore ~\frac{dy}{dx}\\=0-\frac 12+\cos(\log x)\cdot \frac{d}{dx}(\log x) \\=-\frac 12+\frac{\cos(\log x)}{x}~~\text{(ans.)}$

$~4(i)~~(1+x)^x$

Sol. let $~y=(1+x)^x \\ \text{or,}~~\log y= \log(1+x)^x \\ \text{or,}~~ \log y=x \log (1+x) \\ \therefore ~~\text{differentiating w.r.t.}~x, \\~~~\frac 1y\frac{dy}{dx}= \left[\frac{d}{dx}(x)\right]\log(1+x)\\~~~~~~~~~~~~~~~~~~~+x \frac{d}{dx}\left[\log(1+x)\right] \\ \text{or,}~~\frac{dy}{dx}=y\left[1 \cdot \log(1+x)+x \cdot \frac{1}{1+x}\right] \\ \therefore \frac{dy}{dx}=(1+x)^x \left[\log(1+x)+\frac{x}{1+x}\right]~~\text{(ans.)}$

$~(ii)~~x^{\sin x}$

Sol. let $~~y=x^{\sin x} \\ \text{or,}~~ \log y=\log(x^{\sin x}) \\ \text{or,}~~ \log y=\sin x \log x \\ \therefore \frac{d}{dx}( \log y)=\frac{d}{dx}(\sin x) \cdot \log x\\~~~~~~~~~~~~~~~~+ \frac{d}{dx}(\log x)\cdot \sin x \\ \text{or,}~~\frac 1y~\frac{dy}{dx}=\cos x (\log x)+\frac 1x(\sin x) \\ \therefore \frac{dy}{dx}=x^{\sin x}\left[\cos x(\log x)+\frac{\sin x}{x}\right]~~\text{(ans.)}$

$~(iii)~~y=x^{\cos^2x}$

Sol. $~~~y=x^{\cos^2x} \\ \text{or,}~~ \log y=\log(x^{\cos^2x}) \\ \therefore \frac 1y~\frac{dy}{dx}\\=\frac{d}{dx} \left[\cos^2x \log x\right]\\=\frac{d}{dx}(\cos^2x)\cdot \log x+\cos^2x \frac{d}{dx}(\log x)\\=2\cos x(-\sin x) \cdot \log x+\cos^2x \cdot \frac 1x \\ \therefore \frac{dy}{dx}=x^{\cos^2x}\left(-\sin 2x(\log x)+\frac{\cos^2x}{x}\right)~~\text{(ans.)}$

$~(iv)~~x^y=e^x$

Sol.$~~x^y=e^x \\ \text{or,}~~\log_e (x^y)=\log_e(e^x) \\ \therefore ~y \log_e x=x \log_e e \\ \text{or,}~~y \log_e x=x \\ \text{or,}~~y=\frac{x}{\log_e x}\\ \therefore \frac{dy}{dx}\\=\frac{(\log_e x)\frac{d}{dx}(x)-x \cdot \frac{d}{dx}(\log_e x)}{(\log_e x)^2}\\=\frac{(\log_e x) \cdot 1-x \cdot \frac 1x}{(\log_ex)^2}\\=\frac{\log_ex-1}{(\log_ex)^2}\\=\frac{\frac xy-1}{(x/y)^2}~~\left[~\because~\log_ex =\frac xy~\right]\\=\frac{xy-y^2}{x^2}\\=\frac{y(x-y)}{x^2}~~\text{(ans.)}$

$~(v)~~(x^x)^x$

Sol. let $~~y=(x^x)^x=x^{x^2} \\ \text{or,}~~\log y=\log(x^{x^2}) \\ \text{or,}~~ \log y=x^2 \log x \\ \therefore \frac 1y~\frac{dy}{dx}\\=\left[\frac{d}{dx}(x^2)\right](\log x)\\~~~~~~~~~+x^2 \cdot \frac{d}{dx}(\log x)\\=2x \log x+ x^2 \cdot \frac 1x\\=2x \log x+x \\ \therefore \frac{dy}{dx}=(x^x)^x(2x\log x+x)~~\text{(ans.)}$