# Let #f(x)=x^5+3x−2# and let #f^-1# denote the inverse of f, Then how do you find #(f^-1)'(2)#?

##### 1 Answer

#### Explanation:

For the sake of easier notation, we shall say that

According to the definition of inverse functions,

#f(g(x))=x#

Differentiating through the chain rule gives

#f'(g(x))g'(x)=1#

Solving for the derivative of the inverse gives

#g'(x)=1/(f'(g(x)))#

So, we want to find

#g'(2)=1/(f'(g(2))#

We want to first find

Note that if

#f(x)=2" "=>" "x^5+3x-2=2#

Continuing to solve yields

#x^5+3x-4=0#

This may look impossible, but notice that the sum of the coefficients of each term is

Dividing, we see that

#(x-1)(x^4+x^3+x^2+x+4)=0#

Note that

This means that

Returning to what we had earlier, we see that

#g'(2)=1/(f'(g(2)))=1/(f'(1))#

Find this by taking the derivative of

#f(x)=x^5+3x-2" "=>" "f'(x)=5x^4+3#

Thus

#g'(2)=1/8#

With your notation,

#[f^-1]'(2)=1/8#