Question

(a) Differentiate $y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ with respect to $x,0<x<1$,
(b) Differentiate $x^x-2^{\sin x}$ with respect to $x$

Answer 1

$\left(a\right)y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

$\frac{dy}{dx}=\frac{-1}{\sqrt{1-{\left(\frac{1-x^2}{1+x^2}\right)}^2}}\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)$

$=\frac{-(1+x^2)}{\sqrt{{(1+x^2)}^2-{(1-x^2)}^2}}\left[\frac{(1+x^2)(-2x)-(1-x^2)\left(2x\right)}{{(1+x^2)}^2}\right]$

$=\frac{-1}{\sqrt{1+x^4+2x^2-1-x^4+2x^2}}\left[\frac{2x(-1-x^2-1+x^2)}{1+x^2}\right]$

$=\frac{4x^2}{2x(1+x^2)}$

$=\frac{2x}{1+x^2}$

$\left(b\right)y=x^x-2^{\sin x}$

Let $u=x^x$ and $v=2^{\sin x}$

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

We have $u=x^x$ and $v=2^{\sin x}$

$\Rightarrow \log u=\log x^x$ and $\log v=\log 2^{\sin x}$

$\Rightarrow \log u=x\log x$ and $\log v=\sin x\log 2=\log 2\sin x$

$\Rightarrow \frac{1}{u}\frac{du}{dx}=x\times \frac{1}{x}+\log x$ and $\frac{1}{v}\frac{dv}{dx}=\log 2\cos x$

$\Rightarrow \frac{1}{u}\frac{du}{dx}=1+\log x$ and $\frac{1}{v}\frac{dv}{dx}=\log 2\cos x$

$\Rightarrow \frac{du}{dx}=u(1+\log x)$ and $\frac{dv}{dx}=v\left(\log 2\cos x\right)$

$\Rightarrow \frac{du}{dx}=x^x(1+\log x)$ and $\frac{dv}{dx}=2^{\sin x}\left(\log 2\cos x\right)$ where $u=x^x$ and $v=2^{\sin x}$

$\therefore \frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{dx}$

$\frac{dy}{dx}=x^x(1+\log x)-2^{\sin x}\left(\log 2\cos x\right)$