Question

If $y=[2x^3+2][2x^{-3}+1]$, then find $\frac{dy}{dx}$.

Answer 1

Here, $u=2x^3+3,v=2x^{-3}+1$.
Differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}=\frac{d}{dx}\left[\right(2x^3+3\left)\right(2x^{-3}+1\left)\right]$
Using product rule, $\frac{d\left(uv\right)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$
$\frac{dy}{dx}=[2x^3+3]\frac{d}{dx}[2x^{-3}+1]+[2x^{-3}+1]\frac{d}{dx}[2x^3+3]$
$=[2x^3+3][-6x^{-4}]+[2x^{-3}+1]\left[6x^2\right]$
$=[2x^3+2]\left[\frac{-6}{x^4}\right]+[\frac{2}{x^3}+1]\left[6x^2\right]=-\frac{18}{x^4}+6x^3$.