# Thread: FTC to find the derivative of this function

1. ## FTC to find the derivative of this function

I've attached a picture, I have a good handle on using the fundamental theorem of calculus to find derivatives of functions, but this one is just stumping me for some reason. First I tried doing it as is, then I literally had no idea what I could be doing wrong so I tried negating it so that the sin is on top, but that makes no difference. Could someone help a brotha out with this toughie? The thing that's throwing me is the fact that there isn't a real number on the top or bottom of the integral, they're both trig functions. I didn't think this would affect how to go about solving the problem but it must somehow...

2. It's $y'={{\left( 5+{{\cos }^{2}}x \right)}^{10}}(\cos x)'-{{\left( 5+{{\sin }^{2}}x \right)}^{10}}(\sin x)'.$

3. Originally Posted by Krizalid
It's $y'={{\left( 5+{{\cos }^{2}}x \right)}^{10}}(\cos x)'-{{\left( 5+{{\sin }^{2}}x \right)}^{10}}(\sin x)'.$
So basically when you have an integral with two variables, one in each value top and bottom, you subtract the bottom from the top, evaluating it for both, correct?

4. If $f(x)$ is a continuos function in $(g(x),h(x))$ then $F(x)=\int_{g(x)}^{h(x)}f(x)dx\Rightarrow F'(x)=f(h(x))h'(x)-f(g(x))g'(x)$.

So $y'=-(5+\cos^2 x)^{10}\sin x-(5+\sin^2 x)^{10}\cos x$.

5. Gotcha thanks, I wish my professor would cover the finer details of things like this BEFORE assigning us homework involving them

6. But there're some typos there, it should be $
F(x)=\int_{g(x)}^{h(x)}f(t)\,dt\implies F'(x)=f(h(x))h'(x)-f(g(x))g'(x)
$
and in one hand it's $h(x)=\cos x$ not $\cos^2x.$

7. Originally Posted by Krizalid
But there're some typos there, it should be $
F(x)=\int_{g(x)}^{h(x)}f(t)\,dt\implies F'(x)=f(h(x))h'(x)-f(g(x))g'(x)
$
and in one hand it's $h(x)=\cos x$ not $\cos^2x.$