Question

find the derivative of y=integral from -6 to (sq rt)(x) cos t/t^9 dt

Answer 1

You will need to use the first part of the Fundamental Theorem of Calculus (FTC)

Don't forget about the chain rule

Answer 2

hey jim can you help me with my problem once your done here?

Answer 3

sure

Answer 4

ok thanks

Answer 5

I am trying that but can't seem to come up with the correct answer

Answer 6

Using part 1 of the FTC, if

\[\Large g(x) = \int_{a}^{x}f(t)dt \]

then

\[\Large g'(x) = f(x) \]

Answer 7

more generally, if

\[\Large g(x) = \int_{a}^{h(x)}f(t)dt\]

then

\[\Large g'(x) = f(h(x))*h'(x)\]

Answer 8

does this help you get started?