# Find inverse function of exponential function $y=e^x - 2e^{-x}$

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

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.
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 $y=f(x)$, $$(f^{-1})'(y)=\frac{1}{f'(x)}.$$