Finding the derivative of an integral

**Glitch** · October 16th 2010, 02:09 AM

The question:
Find F'(x) for the function F:R -> R given below:
$F(x) = \int^{x^3}_0 {sin(t^2)dt$

My attempt:
It is my understanding that the First Fundamental Theorem of Calculus states that:
F'(x) = f(x)

So my solution was just $sin(x^2)$ . However, this isn't the correct solution. I'm not sure what I'm doing wrong. Any assistance would be appreciated. Thanks.

**skeeter** · October 16th 2010, 02:42 AM

Originally Posted by Glitch

The question:
Find F'(x) for the function F:R -> R given below:
$F(x) = \int^{x^3}_0 {sin(t^2)dt$

My attempt:
It is my understanding that the First Fundamental Theorem of Calculus states that:
F'(x) = f(x)

So my solution was just $sin(x^2)$ . However, this isn't the correct solution. I'm not sure what I'm doing wrong. Any assistance would be appreciated. Thanks.

if $\displaystyle F(x) = \int_a^u f(t) \, dt$ where $u$ is a function of $x$ , then $\displaystyle F'(x) = f(u) \cdot \frac{du}{dx}$

**Glitch** · October 16th 2010, 04:08 AM

Thanks!

**HallsofIvy** · October 16th 2010, 04:27 AM

More generally, Leibniz' rule:
$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x, t)dt= f(x, \alpha(x))\frac{d\alpha}{dx}- f(x, \beta(x))\frac{d\beta}{dx}+ \int_{\alpha(x)}^{\beta(x)} \frac{\partial f}{\partial x}dt$

Which can be derived by the fundamental theorem of calculus and the chain rule as skeeter suggests.

**Glitch** · October 16th 2010, 04:41 AM

Interesting, Leibniz isn't mentioned in this text. Thanks!

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