Let #f(x)=( lne^2x)/x-1# for x1. If g is the inverse of f, then what is g'(3)?

Question

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

Emily Ammerman · Accepted Answer

#g'(3)# is not defined, since #g(3)# does not exist.

First, we can simplify #f(x)#. The logarithm can be split up using the rule: #log(ab)=log(a)+log(b)# #f(x)=(ln(e^2)+ln(x))/x-1# Note that #ln(e^2)=2ln(e)=2#. #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(x)#, and its inverse is arbitrarily named #g(x)#, then we know that #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)#, we see that #g'(x)=1/(f'(g(x)))# So, in order to find #g'(3)#, we will use the equality: #g'(3)=1/(f'(g(3))# The question now becomes, how can we know the value of #g(3)# when we can't even express #g(x)# as a function? Recall that the domain and range of inverse functions are switched, that is, if #g(3)=c#, then #f(c)=3#. So, instead of plugging #x=3# into #g(x)#, which is impossible since we can't find #g(x)#, simply find the #x#-value for which #f(x)=3#. #f(x)=3# #(2+ln(x))/x-1=3# However, graphing this shows that this never occurs. Thus, #g(3)# is not a defined value and neither is #g'(3)#. That was a whole lot of work for nothing! Are you sure you typed the question correctly?

William Abdalla · Answer

undefined To find $ g'(3) $, we first need to find the inverse function $ g(x) $ and then differentiate it to find its derivative.

Given $ f(x) = \frac{\ln(e^{2x})}{x - 1} $ for $ x > 1 $, let's find its inverse $ g(x) $:

$ y = \frac{\ln(e^{2x})}{x - 1} $

Swap $ x $ and $ y $:

$ x = \frac{\ln(e^{2y})}{y - 1} $

Now solve for $ y $:

$ x(y - 1) = \ln(e^{2y}) $

$ xe^{y - 1} = e^{2y} $

$ xe^{y - 1} = (e^y)^2 $

$ \sqrt{x}e^{\frac{y}{2} - \frac{1}{2}} = e^y $

$ \sqrt{x}e^{- \frac{1}{2}} = e^{\frac{y}{2}} $

$ \ln(\sqrt{x}e^{- \frac{1}{2}}) = \frac{y}{2} $

$ 2\ln(\sqrt{x}e^{- \frac{1}{2}}) = y $

$ y = 2\ln(\sqrt{x}e^{- \frac{1}{2}}) $

So, $ g(x) = 2\ln(\sqrt{x}e^{- \frac{1}{2}}) $.

Now, differentiate $ g(x) $ to find $ g'(x) $:

$ g'(x) = \frac{d}{dx}[2\ln(\sqrt{x}e^{- \frac{1}{2}})] $

Using the chain rule and the derivative of $ \ln(u) $:

$ g'(x) = \frac{2}{\sqrt{x}e^{- \frac{1}{2}}} \cdot \frac{1}{2\sqrt{x}} \cdot e^{- \frac{1}{2}} $

$ g'(x) = \frac{1}{x} $

Now, $ g'(3) = \frac{1}{3} $.

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

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