Question

Finding the derivative...

Answer 1

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

Answer 2

oh hey girl

Answer 3

help!!!

Answer 4

I need your help Sapph! :)

Answer 5

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

Answer 6

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

Answer 7

we just need the derivative i believe

Answer 8

I think it's asking about the derivative

Answer 9

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

Answer 10

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

Answer 11

Can you sort of explain this...?

Answer 12

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

Answer 13

does that make sense?

Answer 14

I suck at explaining things :(

Answer 15

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

Answer 16

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

Answer 17

Alright well that makes sense :) Thank you :)

Answer 18

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?

Answer 19

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