Question

HELP!!!!!!!!!!!!
Find the derivative of the following function
F(x)= the integral from sqrtx to 1 of s^2/4+3s^4 ds
I will write it as an equation below as well

Answer 1

\[F(x)=\int\limits_{\sqrt{x}}^{1}\frac{ s^2 }{ 4+3s^4 }ds\]

Answer 2

The FTC1, Fundamental Theorem of Calculus, Part 1 tells us,\[\large \frac{d}{dx}\int\limits_0^x f(t)dt \quad =\quad \frac{d}{dx}\left[F(x)-F(0)\right] \quad = \quad f(x)\]
We'll want to apply this to our problem, but we'll have to be careful, because we also need to apply the chain rule at one point.

Answer 3

\[\large \int\limits\limits_{\sqrt{x}}^{1}\frac{ s^2 }{ 4+3s^4 }ds \qquad = \qquad \int\limits_{\sqrt x}^1 f(s) ds\]
\[\huge = \qquad F(1)-F(\sqrt x)\]
Hmm this might be confusing the way I'm writing it. little f represents the thing we started with, a function of s. big F is what we get when we take the integral.
It's something really messy that we don't want to actually calculate, so we'll leave it like that.

Answer 4

\[\large \frac{d}{dx}F(1)-F(\sqrt x) \quad \rightarrow \quad 0-f(\sqrt x)\frac{d}{dx}\sqrt x\]

Answer 5

Okay could we flip the integral this way \[-\int\limits_{1}^{\sqrt{x}}\frac{ s^2 }{ 4+3s^4}ds\]
and then insert the sqrt x into the spots where there is s?

Answer 6

Yes, that's where we're headed! :)
If the X value is at your lower boundary, then it's being subtracted, meaning we have to attach a negative, as you have done.

But yes, that's a good shortcut, I was just trying to show you some of the fundamentals of it. It's a little confusing though with all the little and big F's :) lol

You can plug the sqrt x into the s's. Just don't forget to multiply the entire thing by the derivative of sqrt x.

Answer 7

Could you help me the rest of the way? I got it wrong

Answer 8

Yah, np :) Sorry I'm working on multiple problems, so my responses might be a little bit slow hehe.

Answer 9

That's alright I appreciate the help

Answer 10

Replacing the s's should give us something like this.\[\huge -\frac{ (\sqrt x)^2 }{ 4+3(\sqrt x)^4}\cdot \frac{d}{dx}(\sqrt x)\]

Answer 11

And we still need to take the derivative of the square root term, hence the d/dx attached to it.

Answer 12

Thanks!

Answer 13

Got through it ok? :) yay team!