## anonymous 5 years ago Finding the derivative...

1. anonymous

h(x)=$\int\limits_{sinx}^{1}\ln(t^2)dt$

2. anonymous

oh hey girl

3. anonymous

help!!!

4. anonymous

I need your help Sapph! :)

5. anonymous

you want the derivative of the integral or the integral itself?

6. anonymous

Umm.. I'm not sure.. What's the difference?

7. anonymous

we just need the derivative i believe

8. anonymous

9. anonymous

Yes, it's the derivative it's asking for.

10. anonymous

using the Fundamental theorem of calculus: $d/dt \int\limits\limits_{\sin x}^{1} \ln (t^2) dt = - \cos x . \ln(\sin^2 x)$

11. anonymous

Can you sort of explain this...?

12. anonymous

ok here is a general formula: $d/dx \int\limits_{g(x)}^{h(x)}f(t) dt = h'(x) f(h(x)) - g'(x) f(g(x))$ I hope that does not complicate it more

13. anonymous

does that make sense?

14. anonymous

I suck at explaining things :(

15. anonymous

Hm.. So it's sort of like the product rule?

16. anonymous

hmm not really.. you just take the derivative of the upper border multiplied by the function after substituting in t minus the same thing with lower border

17. anonymous

Alright well that makes sense :) Thank you :)

18. anonymous

I'm trying to work this out myself.. And well.. I can't seem to get the answer. For the first part I get 0 times f(1)... minus cosx.. How do you get ln?

19. anonymous

it's the same ln in the integral, I just put sinx in the place of t