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How can I find the inverse of

$$y=e^x - 2e^{-x}$$

I normally express the function in terms of $x$. But here, $x$ is in the exponent. If I take $\log$ both sides,

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

Which doesn't appear to help, I still can't get the $x$ out?

The question asks to find $(f^{-1})'(1)$

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You really only need to figure out where the inverse is equal to 1 which is a slightly easier problem than figuring out the entire inverse. In this case that reduces to $1 + 2e^{-x} = e^x$. taking logs gives that $x = ln(1+2e^{-x})$. Rewriting this as $z=ln(x)$ we get $ln(z) = ln(1+\frac{2}{z})$ and in this form it is easy to see that $z=2$ is the solution. Edit: I should add that differentiating y it is easy to see that y is 1-1 so the solution is unique. – Chris Janjigian Feb 5 '12 at 1:56
up vote 8 down vote accepted

As Chris and Isaac noted, you don't actually need to invert the function. But in case you want to do that anyway, the way to do it is to view $$ y = e^x - 2e^{-x} $$ as a quadratic equation in $e^x$ -- namely multiply everything by $e^x$ to get $$ y e^x = (e^x)^2 - 2$$ Now use the quadratic formula to find $e^x$, and then take the logarithm of the result.

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Since you only need the derivative of the inverse function at a single point, you don't need to find a formula for the inverse function. The graph of a function and the graph of its inverse are reflection images of one another over $y=x$, which is equivalent to exchanging $x$ and $y$, so the slope of the tangent lines at corresponding points (and hence the derivative at corresponding points) are reciprocals. If $y=f(x)$, $$(f^{-1})'(y)=\frac{1}{f'(x)}.$$

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