Integration help

**geurrp the yard** · October 13th 2009, 12:51 PM

suppose

, find f'(Pi^(1/3))

Im not sure weather i integrate or diffrenciate.

**redsoxfan325** · October 13th 2009, 01:07 PM

Originally Posted by geurrp the yard

suppose

, find f'(Pi^(1/3))

Im not sure weather i integrate or diffrenciate.

This is an application of the chain rule. Say that $F(y)=\int\sin(5y)\,dy$ . Thus, we are given that $f(t)=F(-t^3)-F(3)$ .

Now take the derivative. (Remember, $F(3)$ is a constant, so its derivative is $0$ .)

$f'(t)=\underbrace{F'(-t^3)=\sin(5\cdot-t^3)\cdot\frac{d}{dt}[-t^3]}_{chain~rule}=\sin(-5t^3)\cdot-3t^2=\boxed{3t^2\sin(5t^3)}$

(Recall that sine is an odd function so $\sin(-5t^3)=-\sin(5t^3)$ .)

**geurrp the yard** · October 13th 2009, 01:12 PM

Hi, Why dont you integrate that sin5y function?

**redsoxfan325** · October 13th 2009, 01:15 PM

You could if you wanted to, but it would be unnecessary work as you would be differentiating again immediately. Also, don't forget to plug $\sqrt[3]{\pi}$ into the boxed equation.

**geurrp the yard** · October 13th 2009, 01:25 PM

Im just a little confused, when I see a integral sign, I automaticaaly assume to integrate. Also why did you not find the derviative of sin?

**redsoxfan325** · October 13th 2009, 01:34 PM

$\frac{d}{dx}\left[\int f(x)\,dx\right]=f(x)$

You don't need to integrate and differentiate to know that's true.

**geurrp the yard** · October 13th 2009, 02:18 PM

Hi when i sub in cube root of Pi into the equation I get zero. Because sin5(pi)Is that correct?

**tom@ballooncalculus** · October 13th 2009, 04:26 PM

Yes I agree

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