Question

A curve has equation $y=(2x-1{)}^{-1}+2x$. $x\ne \frac{1}{2}$
Find $\frac{dy}{dx}$ and $\frac{d^{2}y}{d{x}^2}$.

Answer 1

$y=(2x-1{)}^{-1}+2x$

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

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

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}-2(2x-1{)}^{-2}+2$

$\frac{d^{2}y}{d{x}^2}=8(2x-1{)}^{-3}$.