Question

If $y=f\left(\frac{2x-1}{2x+1}\right)$ and $f^{\prime }\left(x\right)=\sin x^2$, find $\frac{dy}{dx}$

Answer 1

Given $f^{\prime }\left(x\right)=\sin x^2$

$\therefore f^{\prime }\left(\frac{2x-1}{2x+1}\right)=\sin {\left(\frac{2x-1}{2x+1}\right)}^2$

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

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

$=\sin {\left(\frac{2x-1}{2x+1}\right)}^2\times \frac{2(2x+1)-2(2x-1)}{(2x+1{)}^2}$

$\therefore \frac{dy}{dx}=\frac{4\sin {\left(\frac{2x-1}{2x+1}\right)}^2}{(2x+1{)}^2}$