Derivatives of Inverse Functions

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The next session is Sunday, 6:00 - 7:30. It will review the last 2 weeks, then provide time to answer questions about problem sets.

Derivatives of Inverse Functions

Building on the discussion of inverse functions’ derivatives.

Example

How would you find the derivative of y = arcsin x = sin^-1x?

(Beware the “f^-1(x)” notation! It means the inverse of function f, which is not the same as the reciprocal.)

Start by using the fact that any equation of the form y = f^-1(x), for any function’s inverse, means that x = f(y). That lets you rewrite the original equation in terms of a function you don’t know how to differentiate (the inverse sine in this case) into an equation in terms of a function you do know how to differentiate (sine):

Now use implicit differentiation to find dy/dx:

Finally, it would be nice to express this derivative in terms of x, not y — generally derivatives of functions of x are expressed as other functions of x. Fortunately, there’s a trigonometric identity (sin²y + cos²y = 1) that lets you rewrite cosine in terms of sine, and then x, since sin y is x:

The process we followed here to find the derivative of inverse sine is typical of a process you can use to find a derivative of any inverse function:

1. Start with the equation y = f^-1(x); you want to find dy/dx.
2. Rewrite the equation to x = f(y).
3. Use implicit differentiation to find dy/dx.
4. Rewrite occurrences of y in the derivative in terms of x, using already-known relationships between x and y (e.g., y = f^-1(x), x = f(y)) or problem-specific identities.

Another Important Trigonometric Example

What’s the derivative of tan^-1x?

Follow the process outlined above. This time the relevant trigonometric identity is sec²x - tan²x = 1.

Logarithms

What’s the derivative of ln x?

Use the same process as above, remembering that ln (natural logarithm) is the inverse of e^x. The final step, rewriting a derivative involving y in terms of x, is easier than it was for the trigonometric functions, because just setting up the problem provided the necessary relationship between x and y, namely e^y = x:

This result also plugs a hole in one of the antiderivative rules we deduced earlier, namely the antiderivative power rule. That rule doesn’t work for x^-1, because using it would produce division by 0. But now, with the discovery that 1/x (aka x^-1) is the derivative of ln x, we have a way to deal with antiderivatives of x^-1:

Generalization

Can you work out a general formula for the derivative of f^-1(x) in terms of f′(x)?

We can in fact do so, the process we’ve been following for finding derivatives of inverse functions works even in this very general setting, as long as f is differentiable:

But what does this mean, how would you use it on an actual function? Look at how it would work with the natural logarithm function as an example.

Start by cataloging what the f^-1, f, and f′ are:

Then plug them into the formula we derived for the derivative of the inverse function, and simplify:

Next

Look at derivatives of roots (inverses of powers) and try to extend the power rule to fractional powers; collect our new differentiation rules into new antiderivative rules; maybe look at derivatives of logarithms to bases other than e.

There’s no new reading for this, but use the implicit differentiation and inverse functions discussion to explore some of these issues before class time Monday.

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