Question

d/dx * [(integral of (1-t^2) dt from 3 to sin x)]....ill write it nicer with the equation box

Answer 1

\[d/dx ( \int\limits_{3}^{sinx} 1-t ^{2}) dt ) \]

Answer 2

\[\frac{d}{dx}\int\limits_{a}^{g(x)}f(t)dt=f(g(x))g'(x)\]

Answer 3

im not sure how to substitute it into taht equation

Answer 4

\[f(t)=1-t^2\]
\[g(x)=\sin(x)\]
\[a=3\]

Answer 5

does that help?

Answer 6

yes so now it becomes f of sinx*cosx?

Answer 7

\[g'(x)=\cos(x)\]

\[f(g(x))=1-\sin^2(x)=\cos^2(x)\]
\[f(g(x))g'(x)=\cos^3(x)\]

Answer 8

so that is the final answer?

Answer 9

what do you think?

Answer 10

yup

Answer 11

it is

Answer 12

oh cool..thnks for your help! u explain very well