Question

Calculus 2...

Answer 1

Let \[f(x)=\int\limits_{1}^{x}\frac{ dt }{ \sqrt{8+t^4} }\]
Find (f^-1)'(0)

Answer 2

sqrt(8+t^4)

Answer 3

do you know the fundamental theorem of calculus

Answer 4

intergal with limits fx dx = F(b)-F(a)

Answer 5

the derivative of an integral is the integrand

Answer 6

So am I suppose to find the derivative then apply the limits?

Answer 7

do you know the formula to find the derivative of an inverse function?

Answer 8

No

Answer 9

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

Answer 10

so when x = 0, what does the equation look like?

Answer 11

is f(x)^-1 suppose to look like this\[\frac{ 1 }{ \frac{ dt }{ \sqrt{8+t^4}} }\]
Then after that I look for the derivative all under 1?

Answer 12

f'(x) = 1/sqrt(8 + x^4)

Answer 13

now you need to find f^-1(0)

Answer 14

and f^-1(0) = 1, so

1/f'(1)

Answer 15

1/(1/3) = 3. So 3 is your answer