Help with finding the derivatives of invers functions

[clw89]

Hey I need some help with the following problems:

Find (f[sup:7jkbh9ws]-1[/sup:7jkbh9ws])'(a):

1. f(x)= 2x[sup:7jkbh9ws]3[/sup:7jkbh9ws] + 3x[sup:7jkbh9ws]2[/sup:7jkbh9ws] + 7x + 4, a=4

2. f(x)= x[sup:7jkbh9ws]3[/sup:7jkbh9ws] + 3sinx + 2cosx, a=2
i think i did #2 right, but i wanted to make sure. Here is what i did:
2=x[sup:7jkbh9ws]3[/sup:7jkbh9ws] + 3sinx +2cosx
x=0
f[sup:7jkbh9ws]-1[/sup:7jkbh9ws](2)=0
f'=3x[sup:7jkbh9ws]2[/sup:7jkbh9ws] + 3cosx - 2sinx
(f[sup:7jkbh9ws]-1[/sup:7jkbh9ws])'(2)= 1/(3*0 + 3cos0 -2sin0) = 1/3

if i did this right, then there is no point in redoing it, but i'm not sure if i did it right or not.

Thanks for all the help!

 

[tkhunny]

#1

1) f(x) = 2*x^3 + 3*x^2 + 7x + 4
2) f'(x) = 6*x^2 + 6*x + 7 > 0 so there is an inverse and it is nicely behaved.
3) We are interested in a = f(b) = 4, so Solve 2*b^3 + 3*b^2 + 7b + 4 = 4 and determine that b = 0 is the only Real solution.
4) We have the point (0,4) on f(x) and (4,0) on f-inverse(y)

5) \(\displaystyle (f^{-1})^{'}(4)\;=\;\frac{1}{f^{'}(0)}\;=\;\frac{1}{6*(0)^2 + 6*(0) + 7}\;=\;\frac{1}{7}\)

Try to be systematic and deliberate. Good habits learned early WILL save you later.