Derivative of an integral

**angel.white** · December 2nd 2007, 03:54 PM

$h(x)=\int_{6}^{x^{2}} \sqrt{5+r^{3}}dr$

Find h'(x)

Note: I'm having a tough time with these, so explain it like you would to an idiot.

-----------------

This is what I've come up with so far:

$u=x^2$

$\frac{du}{dx}=2x$

$h'(x)=\frac{d}{du}\left(\int_{6}^{u} \sqrt{5+r^{3}}dr\right)\frac{du}{dx}$

$h'(x)=\frac{d}{du}\left(\int_{6}^{u} \sqrt{5+r^{3}}dr\right)2x$

But I can't figure out how to find $\frac{d}{du}\left(\int_{6}^{u} \sqrt{5+r^{3}}dr\right)$

**galactus** · December 2nd 2007, 03:59 PM

Always remember:

$\frac{d}{dx}\int_{h(x)}^{g(x)}f(t)dt=f(g(x))g'(x)-f(h(x))h'(x)$

**angel.white** · December 2nd 2007, 04:18 PM

Thanks, I got
$\sqrt{5+x^{6}}\cdot 2x - \sqrt{5+6^{3}}\cdot 0$

$\sqrt{5+x^{6}}\cdot 2x$

And that was correct. Appreciate it.

**Jhevon** · December 2nd 2007, 07:28 PM

Originally Posted by angel.white

Thanks, I got
$\sqrt{5+x^{6}}\cdot 2x - \sqrt{5+6^{3}}\cdot 0$

$\sqrt{5+x^{6}}\cdot 2x$

And that was correct. Appreciate it.

yes. but note that since the derivative of a constant is zero, you can essentially neglect to write the sqrt(5 + 6^3) part. when you see a constant as a limit, just ignore it.

note that the results here follow by the second fundamental theorem of calculus (as it is sometimes called), galactus' result follows by applying the chain rule to the theorem

**angel.white** · December 2nd 2007, 08:41 PM

Originally Posted by Jhevon

yes. but note that since the derivative of a constant is zero, you can essentially neglect to write the sqrt(5 + 6^3) part. when you see a constant as a limit, just ignore it.

note that the results here follow by the second fundamental theorem of calculus (as it is sometimes called), galactus' result follows by applying the chain rule to the theorem

Thanks

Man, integrals are killing me right now O.o

Got done with all the homework, but I really need to study before the test, but due to my other class schedules and my working, I don't think I'll get another opportunity :/ I might not do as well on this one. Right now I have to just use galactus' method without understanding it, simply b/c of time restraints, but I bought some calc books off amazon.com that I'll try to go through during Christmas break, before Calc2. Then hopefully I will be able to really wrap my head around the integrals.

**Jhevon** · December 2nd 2007, 08:43 PM

Originally Posted by angel.white

Thanks

Man, integrals are killing me right now O.o

Got done with all the homework, but I really need to study before the test, but due to my other class schedules and my working, I don't think I'll get another opportunity :/ I might not do as well on this one. Right now I have to just use galactus' method without understanding it, simply b/c of time restraints, but I bought some calc books off amazon.com that I'll try to go through during Christmas break, before Calc2. Then hopefully I will be able to really wrap my head around the integrals.

look up the second fundamental theorem of calculus. if you don't get it, come back here and ask questions

Thread: Derivative of an integral

LinkBack

Thread Tools

Display

Derivative of an integral

Similar Math Help Forum Discussions

Derivative of an integral

Derivative of integral

Derivative of an Integral

derivative of an integral?

derivative of an integral

Search Tags