Given the function y=f(x), where f(x)=(e^x-e^(-x))/2, find its inverse f'(x).

This is a very common IB-level question,

which is solved with using the technique of a "disguised quadratic".

The first thing we want to do in finding the inverse is swapping the places of x and y in the equation y=f(x), to obtain x=f(y), because finding the inverse is equivalent to reflecting the curve in the line y=x, i.e. interchanging the x and y coordinates. Rearranging the equation x=f(y) to obtain y in terms of x will complete the task.

As such, we have: x=(e^y-e^(-y))/2. We want to solve for y to find the inverse. Firstly, multiply both sides by 2 to obtain: 2x=e^x-e^(-x). Now comes the tricky part: this is actually a quadratic in disguise! We can see this firstly from the rules of exponents: e^(-y)=1/(e^y), so we have: 2x=e^y-1/(e^y). Let us write e^y=p for simplicity: this will render it easier to spot the quadratic. Then 2x=p-1/p, so 2xp=p^2-1, which rearranges to p^2-2xp-1=0. It may not seem like we have done anything special, but remember that p=e^y, so if we can get p in terms of x, we will have e^y in terms of x, so taking a natural logarithm of the eventual expression will gives us the inverse.

Now we can use the quadratic formula to solve for p, where a=1, b=-2x, and c=-1.

We have: p= (2x +- sqrt((-2x)^2+4))/2, which is the same as p= (2x+-sqrt( 4x^2+4))/2. Now factor a 4 out of the expression in the square root to obtain: p= (2x+-sqrt( 4(x^2+1)))/2, or bringing the 4 outside the root: p= x+- sqrt( x^2+1).

Recall that p=e^y so we wouldn't want the expression for p to be negative, since we then wouldn't be able to take the logarithm of a negative value. As such, we take the positive square root in the expression for p: p=x + sqrt(x^2+1). So e^y= x + sqrt(x^2+1), or y= ln(x+sqrt(x^2+1)).

And there we have it: the inverse of y=f(x), where f(x)=(e^x-e^(-x))/2, is y= ln(x+sqrt(x^2+1)).