Question

Differentiate: $({\sin }^{-1}x+\frac{1}{2}\log \frac{1+x}{1-x})$.

Answer 1

$\frac{d}{dx}({\sin }^{-1}x+\frac{1}{2}\log \frac{1+x}{1-x})$

$=\frac{d}{dx}({\sin }^{-1}x+\frac{1}{2}(\log 1+x-\log (1-x))$ $[\log \frac{b}{a}=\log b-\log c]$

$=\frac{d}{dx}{\sin }^{-1}x+\frac{d}{dx}\frac{\log 1+x}{2}-\frac{d}{dx}\frac{\log 1-x}{2}$

$=\frac{d}{dx}{\sin }^{-1}x+\frac{d}{dx}\frac{\log 1+x}{2}-\frac{d}{dx}\frac{\log 1-x}{2}$

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

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

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

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

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

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