Question

What is the derivative of y=(x^2-1)^3

Answer 1

Do you know how to use the chain rule?
y = f(g(x))
y' = f'(g(x)) * g'(x)

Answer 2

somewhat

Answer 3

\[\frac {d}{dx} u^n = n u^{n-1} \frac {du}{dx}\]\[\frac {d}{dx} \left[ y=(x^2-1)^3 \right]\]\[\frac {dy}{dx} = 3(x^2 - 1)^2 \times \frac {d}{dx} (x^2 - 1)\]\[\frac {dy}{dx} = 6x(x^2 - 1)^2\]

Answer 4

Yes, that would be correct for the derivative. :P

If we look at the f(x) to be x^3 and the g(x) to be (x^2 - 1), we'd have
f(g(x)) = f(x^2 - 1) = (x^2 - 1)^3

The derivative of f(x) is 3x^2
f'(x) = 3x^2
f'(g(x)) = 3(x^2 - 1)^2

The derivative of g(x) would be 2x

If we use that "y' = f'(g(x)) * g'(x)", we'd just insert those back in...

y' = 3(x^2 - 1)^2 * (2x)
= 6x(x^2 - 1)^2