Answer:
[tex]\frac{dy}{dx}[/tex] = - [tex]\frac{18x}{(3x^2+1)^4}[/tex]
Step-by-step explanation:
Differentiate using the chain rule
Given
y = f(g(x)) , then
[tex]\frac{dy}{dx}[/tex] = f'(g(x) × g'(x)
Given
y = [tex]\frac{1}{(3x^2+1)^3}[/tex] = [tex](3x^2+1)^{-3}[/tex] , then
[tex]\frac{dy}{dx}[/tex] = - 3[tex](3x^2+1)^{-4}[/tex] × [tex]\frac{d}{dx}[/tex] (3x² + 1 )
= - 3[tex](3x^2+1)^{-4}[/tex] × 6x
= - [tex]\frac{18x}{(3x^2+1)^{4} }[/tex]
