Question

Hi. I need help with this question (see image). Please show workings.​

Answer 1

Answer:

$\frac{dy}{dx}$ = - $\frac{18x}{(3x^2+1)^4}$

Step-by-step explanation:

Differentiate using the chain rule

Given

y = f(g(x)) , then

$\frac{dy}{dx}$ = f'(g(x) × g'(x)

Given

y = $\frac{1}{(3x^2+1)^3}$ = $(3x^2+1)^{-3}$ , then

$\frac{dy}{dx}$ = - 3$(3x^2+1)^{-4}$ × $\frac{d}{dx}$ (3x² + 1 )

    = - 3$(3x^2+1)^{-4}$ × 6x

    = - $\frac{18x}{(3x^2+1)^{4} }$