2 questions about derivatives of integrals?

Question

I really need help on this, exam is tomorrow morning.
If F(x)=Integral from 0 to sqrt(1+x^2) of cos(t^2)dt. To what is equal its derivative? I tried this way : I put G(y)=integral from 0 to y of cos(t^2)dt. I notice that F(x)=G(sqrt(1+x^2)), so F'(x)=G'(sqrt(1+x^2)).2x/(2sqrt(1+x^2)). At final it gave me xcos(1+x^2)/sqrt(1+x^2). Is it right?
Then come the big problem. I could do this only because I knew how to derivate an integral starting from 0 to something else, but now, I have to derivate the integral from cosx to x^4 of 1/(1+z^2)dz. How to do so? Please explain each step and give me the answer.

Answer 1

For the first part: Yes, that's correct.

For the second part: Actually, this is a lot easier than you think. Note that ∫(a to b) is the same as ∫(0 to b) - ∫(0 to a). So if we have F(x) = ∫(g(x) to h(x)) f(z) dz, then we can adopt the same procedure: let G(y) = ∫(0 to y) f(z) dz, then F(x) = G(h(x)) - G(g(x)), and chain rule says
F'(x) = G'(h(x)).h'(x) - G'(g(x)).g'(x)
= f(h(x)).h'(x) - f(g(x)).g'(x)
So in this case we have
[1 / (1 + (x^4)^2)] . 4x^3 - [1 / (1 + (cos x)^2)] . (-sin x)
= 4x^3 / (1 + x^8) + sin x / (1 + cos^2 x).