# Somewhat confused on the fundamental theorem of calculus

I have the following equation which I require to find the derivative by using the Fundamental theorem of calculus: $$f(x) = \int_3^{x^2} \left(\frac13t^2-1\right)^{15}dt$$

Trying to understand this, I tried plugging $f(x^2)$ in the formula minus $f(3)$ which was wrong.

Originally, I thought it be $F(x^2)-F(3)$ but I cannot seem to find the anti-derivative. Can someone help pls ? It's been over a couple years, so I do not remember my derivatives rules by heart, please be clear about those.

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Hint: If $$f(x)=\int_3^{x^2}g(t)dt,$$ then $$f'(x)=g(x^2)\cdot 2x,$$ where the $2x$ is from the derivative of $x^2$.

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@Reza M.: Google Leibniz rule to find out what Clayton did. +1 –  Babak S. Jan 28 '13 at 6:33
!!!Thanks, I was doing the anti-derivative of $x^2$ >< Thanks –  Reza M. Jan 28 '13 at 6:34
@RezaM.: You're welcome. –  Clayton Jan 28 '13 at 6:35

The great thing about the fundamental theorem of calculus is that you don't need to find the anti-derivative in order to differentiate the integral. Write

$$f(x)=F(x^2)-F(3)$$

Then, since $F^\prime=(\frac{1}{3}t^2-1)^{15}$,

$$\frac{df}{dx}=\frac{d}{dx}(F(x^2)-F(3))=\frac{d}{dx}F(x^2)=F^\prime(x^2)\cdot 2x=2x\left(\frac{1}{3}x^4-1\right)^{15}$$

Notice that $\frac{d}{dx}F(3)=0$ since $F(3)$ will be a constant.

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!!!Thanks, I was doing the anti-derivative of $x^2$ >< Thanks –  Reza M. Jan 28 '13 at 6:39