anonymous 3 years ago solving for g'(x) when g(x)=integral from sqrt(x) to x^2, (sin t)(sqrt t) dt

1. anonymous

derivative of the integral is the integrand, plus chain rule replace $$t$$ by $$x^2$$ and multiply by the derivative of $$x^2$$ get $\sin(x^2)\sqrt{x^2}\times 2x$ for the top part bottom part is the same, but make it negative

2. anonymous

I think I am suppose to break it up. integral from sqrt x to 0 plus 0 to x^2. But i have no idea why this is the way to do it

3. anonymous

yes but you can break up anyway you like

4. anonymous

the top part gives what i wrote above, the bottom part is analogous, just make it negative

5. anonymous

the bottom can be any number??

6. anonymous

I know I'm suppose to use the second fundamental theorem

7. anonymous

yes $\int_{\sqrt{x}}^{x^2}f(t)dt=\int_{\sqrt{x}}^af(t)dt+\int_a^{x^2}f(t)dt$

8. anonymous

My teacher chose 0 though. Is there a reason for that? should I just choose 0 everytime?

9. anonymous

take the derivative of the second part, get $$f(x^2)\times 2x$$

10. anonymous

the derivative of the first part is $$-f(\sqrt{x})\times \frac{1}{2\sqrt{x}}$$

11. anonymous

if you like to pick zero, go ahead a pick zero makes no difference

12. anonymous

Ok I understand it now. Thanks!!!! :)