Finding the derivative...

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Finding the derivative...

Mathematics
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h(x)=\[\int\limits_{sinx}^{1}\ln(t^2)dt\]
oh hey girl
help!!!

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I need your help Sapph! :)
you want the derivative of the integral or the integral itself?
Umm.. I'm not sure.. What's the difference?
we just need the derivative i believe
I think it's asking about the derivative
Yes, it's the derivative it's asking for.
using the Fundamental theorem of calculus: \[d/dt \int\limits\limits_{\sin x}^{1} \ln (t^2) dt = - \cos x . \ln(\sin^2 x)\]
Can you sort of explain this...?
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
does that make sense?
I suck at explaining things :(
Hm.. So it's sort of like the product rule?
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
Alright well that makes sense :) Thank you :)
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?
it's the same ln in the integral, I just put sinx in the place of t

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