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

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?

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

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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} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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