Question

For f(x)=integral 1/sqrt(4+e^t)dt from 0 to lnx, with x>0, find (f^-1)'(0)

Answer 1

so is it \[f(x)=\int_0^{\ln x}\dfrac{dt}{4+e^t},x>0\\f^{-1}(0)=?\]

Answer 2

or so is it \[f{'}(0)=?\]

Answer 3

it is (f^-1)'(0)= It's asking for the derivative of the inverse of f at 0

Answer 4

correct

Answer 5

it's asking for the derivative of f^-1 at x = 0

Answer 6

yes, that is correct

Answer 7

do you know the formula?

Answer 8

I'm not sure what formula to use

Answer 9

\[\frac{d}{dx} f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}\]

Answer 10

two things you need to find
d/dx f(x)

and f^-1(0)

Answer 11

I believe the derivative is 1/xsqrt(4+x). Now I have to find the inverse at 0?

Answer 12

f^-1(0) means
0 = f(x)

Answer 13

do you know how to find 0 = f(x) ?

Answer 14

would i just put 1/sqrt(4+e^t)=0?

Answer 15

no, its