## justinfung Group Title solving for g'(x) when g(x)=integral from sqrt(x) to x^2, (sin t)(sqrt t) dt one year ago one year ago

1. satellite73 Group Title

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. justinfung Group Title

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. satellite73 Group Title

yes but you can break up anyway you like

4. satellite73 Group Title

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

5. justinfung Group Title

the bottom can be any number??

6. justinfung Group Title

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

7. satellite73 Group Title

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

8. justinfung Group Title

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

9. satellite73 Group Title

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

10. satellite73 Group Title

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

11. satellite73 Group Title

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

12. justinfung Group Title

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