Recall that the most important property of an inverse function is that it *undoes* the action of a function, and vice versa. In function notation it looks like this:

It might help to think of a function as an "**operator**" which **operates** on the variable **x**. If the function **f** operates on **x**, then operation by the inverse function **f ^{-1}** immediately afterward simply

It works the other way too. Operation by **f** on the result of operating on **x** by **f ^{-1}** just undoes the first action. Inverse operations undo each other.

We recognize the importance of inverse operations when we work with trigonometric and exponential functions. We can't find an angle in a right triangle from the lengths of its sides without an inverse trig. function (sin** ^{-1}**(x), cos

**Inverse functions undo the action of each other**

Graphically, a function and its inverse are mirror images across the line **y = x**. Take the example plotted below. The inverse of **f(x) = x ^{2}** is the square root function,

The blowup on the right gives you a closer view. Now the way that reflection across the line **y = x** works is kind of cool. If a point **(x, y)** exists on **f(x)**, then the point **(y, x)** is its mirror image on **f ^{-1}(x)**. For example,

This mirror-image property will help us a lot as we take derivatives of inverse functions.

The graphs of a function and its inverse are mirror images across the line **y = x**.

Now let's take a generic function, **f(x)**, and its inverse **f ^{-1}(x)**. We begin with the statement

Now we differentiate both sides with respect to **x**,

remembering that we need to apply the chain rule to the composition of functions on the left:

If we divide to isolate the derivative of the inverse on the left, we finally get:

The derivative of an inverse function, **f ^{-1}(x)** can be found without directly taking the derivative, if we know the function,

Finding the derivatives of the main inverse trig functions (sine, cosine, tangent) is pretty much the same, but we'll work through them all here just for drill

We're looking for

If we let

then we can apply **f(x) = sin(x)** to both sides to get:

Now if we take the derivative of each side with respect to **x** (**d/dx**), remembering that we have to use implicit differentiation on the left, we get:

On the right is just = 1, so we have the derivative we're looking for:

... but we want it in terms of **x**. That's easy because we know that **sin(y) = x**. We can build a triangle that reflects that fact: the sine of angle **y** is **x**:

To get the bottom side, just use the Pythagorean theorem. Now we can replace **cos(y)** with an algebraic expression containing **x**:

First let **y = tan ^{-1}(x)**, then apply

Now taking the derivative of both sides gives:

Then we divide by **sec ^{2}(x)** to get

Now using **tan(y) = x**, we can construct another triangle:

... and from that we find

This derivative is calculated in much the same way. We'll skip the details for this one; you should try it on your own. The result is:

You could use the same method to find derivatives of the inverse cosecant, secant and cotangent functions, too. Try it!

Solution: We begin with our function and its inverse:

We need the derivative of the function

Then it's just a matter of plugging the inverse in to cos(x):

Now it's a little difficult to convert this into the form we found in example 1, but if we plot the two, the result is in the graph on the right.

I've shifted the purple curve upward by 0.1 so you can see that the curves would perfectly overlap otherwise.

This graph is another version of the **f(x) = x ^{2}** ,

Let's say we want to know the derivative (slope) of the inverse function at x = 4, but we don't actually know the inverse function (I know we know it here, but pretend we don't). Turns out we don't really need to know f-1(x).

If (4, 2) is a point on f-1(x), then (2, 4) is the point on f(x) at which f(x) has the reciprocal slope.

The green lines are tangent to the functions at those points. They have reciprocal slope. That means that the slope of the inverse at x = 4 can be found by taking the derivative of **f(x)** at x = 2.

We saw above that if points **(a, b)** and **(c, d)** lie on **f(x)**, then points **(b, a)** and **(d, c)** will lie on **f ^{-1}(x)**. We can use those two points to calculate the slopes of the segments connecting those lines. On the function, that slope is

and on the inverse it is.

It's easy to recognize that these are reciprocals of each other:

locations **a**, **b**, **c** and **d** are completely arbitrary, so we could make the distance between our two points zero, as we would in taking a derivative, so the relationship between derivatives is:

1. Let . Find at x = 1.

2. If **f(x) = x ^{3} + 7x + 2**, and

3. If **f(x) = 7x ^{3} + (ln x)^{3}**, and

4. Find the equation of the line tangent to the inverse of **f(x) = x ^{5} + 2x^{3} + x - 4** at the point

5. Take a look at the table below, showing values of a function **f(x)** and its derivative, **f'(x)**. Use the table to find **(f ^{-1})'(1)** and

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