anonymous
  • anonymous
Finding the derivative...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
h(x)=\[\int\limits_{sinx}^{1}\ln(t^2)dt\]
anonymous
  • anonymous
oh hey girl
anonymous
  • anonymous
help!!!

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

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