Question

How do you use the chain rule to differentiate `y=1/(2x^3+7)`?

Answer 1

the derivative is `dy/dx=(-6x^2)/((2x^2+7)^2)`

Explanation:

Let `u(x)=2x^3+7`

then `y=(u(x))^(-1)`

so the derivative is `dy/dx=-1(u(x))^(-2)*u'(x)`

we use the following `(x^n)'=nx^(n-1)`

and `u'(x)=(2x^3+7)'=6x^2`

So the final result is `(-1/(2x^3+7)^2)*6x^2`