Question

If $y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$, find the value of $\frac{dy}{dx}$.

Answer 1

Let $y=si{n}^{-1}\left(\frac{2x}{1+x^2}\right)$

We need to find $\frac{dy}{dx}$.

Let $u=\frac{2x}{1+x^2},f=si{n}^{-1}\left(u\right)$

Apply chain rule: $\frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}$

$\therefore \frac{df\left(u\right)}{dx}=\frac{d}{du}\left(si{n}^{-1}\right(u\left)\right)\cdot \frac{d}{dx}\left(\frac{2x}{1+x^2}\right)$

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

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

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

$\therefore \frac{df\left(u\right)}{dx}=\frac{1}{\sqrt{1-u^2}}\cdot \frac{2(1-x^2)}{(1+x^2{)}^2}$

Substitute back $u=\frac{2x}{1+x^2}$ we get

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

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

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

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

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

$\therefore \frac{dy}{dx}=\frac{2(1-x)(1+x)}{(x+1)(x-1)(1+x^2)}$

$\therefore \frac{dy}{dx}=\frac{2}{(1+x^2)}$