Put $f(x) =\int_0^{t^2} \sin(x^4) dx$ and give a formula for $f'(t)$ $$f\left(t\right) = \int_{0}^{t^2} \sin\left(x^4\right)\ {\rm d}x$$
My answer:
The chain rule is
$$g'(t)f'(g(t))$$
So if
$$g(t) = t^2$$ and
$$f'(x) = \sin(x^4)$$
Then $f'(t)$ must be
$$f'(t) = 2t \sin(t^4)$$
Which was wrong. What is the correct way to approach this problem?

Hint: Another method is to try to find a change of variables so that you can write
$$f(t) = \int_c^t g(x)\; dx$$
and then you have $f'(t) = g(t)$. The important thing is that you end up with "$t$" as the upper limit.
