Finding the derivative...

Question

anonymous · Answer

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

anonymous · Answer

oh hey girl

anonymous · Answer

help!!!

anonymous · Answer

I need your help Sapph! :)

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

we just need the derivative i believe

anonymous · Answer

I think it's asking about the derivative

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

Can you sort of explain this...?

anonymous · Answer

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

anonymous · Answer

does that make sense?

anonymous · Answer

I suck at explaining things :(

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

Alright well that makes sense :) Thank you :)

anonymous · Answer

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?

anonymous · Answer

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