# How to find derivative $f'(t)$ using the integral from $0$ to $t^2$ of $\sin(x^4)$?

If the $t^2$ was just $t$, I know the derivative is $\sin(t^4)$. How does having an upper limit of $t^2$ change $\sin(x^4)$?

Thank you.

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You need the chain rule to help. See this $$\left(\int_a^{g(x)} f(t)\,dt = \right)'= f\circ g(x)g'(x)$$

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Is this saying that the derivative is g(x) * g'(x)? I have done chain rule for computing derivatives but not when integrals are involved. Thanks –  koloa Mar 10 '13 at 2:08
This is the chain rule along with the fundamental theorem of calculus (derivative of integral version) –  ncmathsadist Mar 10 '13 at 2:18