Inverse function + integration help

[theneedtoknow]

Homework Statement

find (f-1)'(0) if f'(x) = root(1+x^4)

The Attempt at a Solution

i know (f-1)'(0) = 1 / f'((f-1 (0))

but to find the value of the inverse at 0, i need to find the inverse, for which i need to find the original function by integrating, and i cannot seem to be able to integrate root (1+x^4)

i mean..if I set u to 1+x^4, i get du = 4x^3 dx
but there is nothing other than what's under the root sign in the integral so it doesn't work out , so please guide me :)


and another question
f(x) = (ax + b) / (cx + d)
Determine a, b , c, d for which f = f-1

so i foudn the inverse to be f(x) = (dx - b) / (a - cx)

so i set the two functions equal to each other

(ax + b) / ( cx + d) = (dx - b) / (a - cx)
I have no idea how to figure out the constants now though...After some algebra i get
cx^2 (d+a)+ x(d+a)(d-a) - b(d+a) = 0
so assuming d+ a is not equal to zero, i can simplify to
cx2 +x(d-a)-b = 0
but how the hell do i figre out the constants from this point?

 

[rootX]

There is a rule that says
derivative of inverse = 1/derv of the function

e.g.
y = x^2
dy/dx = 2x

x = sqrt(y)
dx/dy= -1/(2*sqrt(y))

oo nvm .. perhaps, it was something else.

but, you can develop the right thing yourself:

f(f^-1) = x
differentiate and use chain rule. Hopefully that would give you an answerfor b) I think that is much more like common sense question. How you draw an inverse function?!

Have a graph paper and draw first a function and then its inverse, and then that would answer your question