Differentiation (Part-38) | S N De

$~4(vi)~~(\tan x)^{\sin x}~~\text{when}~~x=\frac{\pi}{4}$

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

For $\,x=\frac{\pi}{4},\,$ 

the value of $~~y=\left(\tan \frac{\pi}{4}\right)^{\sin(\pi/4)}\\~~~~~=(1)^{1/\sqrt 2}=1$

Hence, from $\,(1)\,$ we get,

$~~~~\left[\frac{dy}{dx}\right]_{x=\frac{\pi}{4}}\\=1 \times [\cos(\pi/4) \log(\tan(\pi/4))\\~~~~~~~+\sec(\pi/4)]\\=\frac{1}{\sqrt 2} \times \log 1+\sqrt 2\\=\sqrt 2~~\text{(ans.)}$

$~(vii)~~x^{\cos^{-1}x}$

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

$~(viii)~~x^{e^x}$

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

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$~(ix)~~(\log x)^{\cos x}$

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

$~(x)~~10^x \cdot x^{10}$

Sol. let $~~y=10^x \cdot x^{10} \\ \text{or,}~~ \log_e y=\log_e(10^x \cdot x^{10}) \\ \text{or,}~~ \log_e y=\log_e 10^x + \log_e x^{10} \\ \text{or,}~~ \log_e y=x \log_e 10+10 \log_e x \\~~~~~\text{Differentiating w.r.t.}~x, \\~~~\frac 1y~\frac{dy}{dx}=1 \cdot \log_e 10+10 \cdot \frac 1x \\ \therefore ~\frac{dy}{dx}=\frac yx(x\log_e10+10)  \\ \text{or,}~~ \frac{dy}{dx}=\frac{10^x \cdot x^{10}}{x}\left(x \log_e 10+10\right) \\ \text{or,}~~ \frac{dy}{dx}=10^x \cdot x^9(x\log_e10+10)~~\text{(ans.)}$

$~(xi)~~e^x\tan x$

Sol. $~~y=e^x \tan x \\ \text{or,}~~ \frac{dy}{dx}=\frac{d}{dx}(e^x \tan x) \\ \text{or,}~~  \frac{dy}{dx}=\frac{d}{dx}(e^x) \cdot \tan x+e^x \cdot \frac{d}{dx}(\tan x) \\ \text{or,}~~  \frac{dy}{dx}=e^x \tan x+e^x \sec^2x \\ \therefore~~\frac{dy}{dx}=e^x(\tan x+\sec^2x)~~\text{(ans.)}$

$~(xii)~~x^3 \log x$

Sol. let $~~y=x^3 \log x \\ \therefore \frac{dy}{dx}=\frac{d}{dx}(x^3\log x) \\ \text{or,}~~ \frac{dy}{dx}=\frac{d}{dx}(x^3) \cdot \log x+x^3 \frac{d}{dx}(\log x) \\ \text{or,}~~ \frac{dy}{dx}= 3x^2 \log x+x^3 \cdot \frac 1x \\ \text{or,}~~ \frac{dy}{dx}=3x^2 \log x+ x^2 \\ \text{or,}~~ \frac{dy}{dx}=x^2(3 \log x+1)~~\text{(ans.)}$

$~(xiii)~~\sqrt x \log \sqrt x$

Sol. let $~~y=\sqrt x\log \sqrt x \\ \text{or,}~~\frac{dy}{dx}=\frac{d}{dx}(\sqrt x \cdot \frac 12 \log x)\\ \text{or,}~~ \frac{dy}{dx}=\frac{d}{dx}(x^{1/2}) \cdot \frac 12 \log x\\~~~~~~~~~+\sqrt x~\frac{d}{dx}(\frac 12 \log x) \\ \text{or,}~~\frac{dy}{dx}=\frac 12 x^{1/2-1} \cdot \frac 12 \log x\\~~~~~~~~~+\frac 12 \sqrt x \cdot \frac 1x \\ \text{or,}~~ \frac{dy}{dx}=\frac{1}{2\sqrt x} ~\log x^{1/2}+\frac{1}{2\sqrt x}\\ \text{or,}~~ \frac{dy}{dx}=\frac{1}{2\sqrt x}(\log  \sqrt x+1) ~~\text{(ans.)}$

$~(xiv)~~e^x\sec x$

Sol. let $~~y=e^x \sec x \\ \therefore \frac{dy}{dx} =\frac{d}{dx}(e^x \sec x) \\ \text{or,}~~ \frac{dy}{dx}=\frac{d}{dx}(e^x)\cdot \sec x\\~~~~~~~~~~~~+ e^x \frac{d}{dx}(\sec x) \\ \text{or,}~~ \frac{dy}{dx}=e^x \sec x+e^x \sec x\tan x \\ \text{or,}~~ \frac{dy}{dx}=e^x \sec x(1+\tan x)~~\text{(ans.)}$

$~(xv)~~x \cdot \frac{e^x+e^{3x}}{e^x+e^{-x}}$

Sol. $~~y=x \cdot \frac{e^x+e^{3x}}{e^x+e^{-x}} \\ \text{or,}~~y=x\cdot \frac{e^x(e^x+e^{3x})}{e^x(e^x+e^{-x})}   \\ \text{or,}~~ y= x~ \cdot \frac{e^{2x}+e^{4x}}{e^{2x}+1}  \\ \text{or,}~~ y=x ~\cdot \frac{e^{2x}(1+e^{2x})}{(1+e^{2x})}  \\ \text{or,}~~ y=x ~\cdot e^{2x} \\ \therefore ~~\frac{dy}{dx}=\frac{d}{dx}(x \cdot e^{2x})  \\ \text{or,}~~ \frac{dy}{dx}=\frac{d}{dx}(x) \cdot e^{2x}+x \cdot \frac{d}{dx}(e^{2x})  \\ \text{or,}~~ \frac{dy}{dx}= 1 \cdot e^{2x}+ x ~\cdot 2e^{2x}  \\ \text{or,}~~ \frac{dy}{dx}=e^{2x}(1+2x)~~\text{(ans.)}$

$~(xvi)~~x\sec x\log(xe^x)$

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