# Let #f(x)=( lne^2x)/x-1# for x>1. If g is the inverse of f, then what is g'(3)?

##### 1 Answer

#### Explanation:

First, we can simplify

The logarithm can be split up using the rule:

#f(x)=(ln(e^2)+ln(x))/x-1#

Note that

#f(x)=(2+ln(x))/x-1#

We can try to take the inverse of this function, but it will not work very well. So, the question becomes, how can we find the value of the function's inverse's derivative?

Let's backtrack and come up with a rule regarding inverse functions and derivatives. If we have a function

#f(g(x))=x#

This is the definition of inverse functions. However, if we differentiate both sides, then we see that:

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

Note the use of the chain rule on the left-hand side. Solving for the derivative of the inverse function,

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

So, in order to find

#g'(3)=1/(f'(g(3))#

The question now becomes, how can we know the value of

Recall that the domain and range of inverse functions are switched, that is, if

#f(x)=3#

#(2+ln(x))/x-1=3#

However, graphing this shows that this never occurs. Thus,

That was a whole lot of work for nothing! Are you sure you typed the question correctly?