Derivative of inverse functions

When we do a composition of a function and its inverse the result is that we do some operation, undo it with the reversed operation, which results in the output and the input being equal to each other. In mathematical notation: [math]\displaystyle{ f(f^{-1}(x)) = x }[/math]. For now we skip the conditions for which a function is invertible. To make the proof easier to read let's write [math]\displaystyle{ f^{-1}(x) = g(x) }[/math]:

The rate of change of [math]\displaystyle{ x }[/math] is trivial, it's 1. If [math]\displaystyle{ \frac{d}{dx}x = 1 }[/math], then [math]\displaystyle{ \frac{d}{dx}f(g(x)) = g'(x)f'(g(x)) }[/math] by the chain rule.

Now we have this (it's the concept of implicit differentiation):

[math]\displaystyle{ g'(x)f'(g(x)) = 1 }[/math]

Which can be rewritten as

[math]\displaystyle{ g'(x) = \frac{1}{f'(g(x))} }[/math]

Careful! [math]\displaystyle{ f'(g(x)) \neq [f(g(x))]' }[/math]. Else you'll apply the chain rule twice.

What we have proven is that if we know [math]\displaystyle{ f' }[/math], then we can find [math]\displaystyle{ g' }[/math] without having to explicitly calculate it. This property can be used to prove the derivative of the exponential without having to use limits of sums. We can also do the same for the inverse trigonometric functions, because if we try to use the definition, we are going to have some very difficult limits to solve.

In case we have to differentiate a function but we can't solve the limit because it requires some property that we don't know, we can rely on its inverse because it may be a function for which the derivative is known. We are applying the concept of implicit differentiation because we are expressing a derivative in terms of some other derivative which we already know.