Question

Differentiate the given functions w.r.t. x.

$(sinx{)}^x+si{n}^{-1}\sqrt{x}.$

Answer 1

y = $(sinx{)}^x+si{n}^{-1}\sqrt{x}$

Let $u=(sinx{)}^x,v=si{n}^{-1}\sqrt{x}$

$\therefore$ y = u + v

Differentiating w.r.t., x

$\Rightarrow \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

Now, $u=(sinx{)}^x$

Taking log on both sides, $logu=log(sinx{)}^x\Rightarrow logu=xlog(sinx)$

Differentiating w.r.t., x, we get

$\frac{1}{u}\frac{du}{dx}=x\frac{d}{dx}log\left(sinx\right)+log\left(sinx\right)\frac{d}{dx}\left(x\right)\left(Usingproductrule\right)=\frac{x}{sinx}cosx+log\left(sinx\right)\frac{du}{dx}=u[xcotx+logsinx]=\left(sinx{)}^x\right[xcotx+logsinx]Again,v=si{n}^{-1}\sqrt{x}Differentiatingw.r.tx,wegetNow,v=si{n}^{-1}\sqrt{x}\frac{dv}{dx}=\frac{1}{\sqrt{1-(\sqrt{x}{)}^2}}\frac{d}{dx}{x}^{1/2}=\frac{1}{\sqrt{1-x}}\frac{1}{2}{x}^{-1/2}(Usingchainrule)\Rightarrow \frac{dv}{dx}=\frac{1}{\sqrt{1-x}}\times \frac{1}{2\sqrt{x}}\Rightarrow \frac{1}{2\sqrt{x}\sqrt{1-x}}\Rightarrow \frac{dv}{dx}=\frac{1}{2\sqrt{x-x^2}}Puttingthevaluesof\frac{du}{dx}and\frac{dv}{dx}inEq.(i),weget\frac{dy}{dx}=(sinx{)}^x[xcotx+logsinx]+\frac{1}{2\sqrt{x-x^2}}$