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:
$$f(f^{1}(x)) = f^{1}(f(x)) = x$$
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 undoes that action.
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^{1}(x), ... ), and we can't solve for a variable locked in an exponent unless we can take a log, the inverse function of exponentiation.
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, f^{1}(x) = √x. Notice that for the root function, we have to restrict ourselves to the upper arm of the sideways parabola, otherwise it would be doublevalued. That's very common with inverse functions (see Inverse Trig. Functions).
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, (2, 4) is on the graph of f(x) below, and (4, 2) is its mirror image on f^{1}(x).
This mirrorimage 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. If (x, y) is on f(x), then (y, x) is its mirrorimage point across y = x, and the slope of f(x) at x is the reciprocal of the slope of f^{1}(x) at y.
Now let's take a generic function, f(x), and its inverse f^{1}(x). We begin with the statement
$$f(f^{1}(x)) = x$$
Now we differentiate both sides with respect to x,
$$\frac{d}{dx} f(f^{1}(x)) = \frac{d}{dx}x$$
remembering that we need to apply the chain rule to the composition of functions on the left:
$$f'(f^{1}(x))[f^{1}(x)]' = 1$$
If we divide to isolate the derivative of the inverse on the left, we finally get:
$$[f^{1}(x)]' = \frac{1}{f'(f^{1}(x))}$$
$$\frac{d}{dx} f^{1}(x) = \frac{1}{f'(f^{1}(x))}$$
The derivative of an inverse function, f^{1}(x) can be found without directly taking the derivative, if we know the function, f(x), and its derivative.
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
$$\frac{d}{dx} sin^{1}(x)$$
If we let
$$y = sin^{1}(x)$$
then we can apply f(x) = sin(x) to both sides to get:
$$sin(y) = x$$
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:
$$cos(y) \frac{dy}{dx} = \frac{dx}{dx} = 1$$
On the right is just = 1, so we have the derivative we're looking for:
$$\frac{dy}{dx} = \frac{1}{cos(y)}$$
... 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:
$$$$
To get the bottom side, just use the Pythagorean theorem. Now we can replace cos(y) with an algebraic expression containing x:
$$\frac{d}{dx} sin^{1}(x) = \frac{1}{\sqrt{1  x^2}}$$
$$\frac{d}{dx} tan^{1}(x)$$
First let y = tan^{1}(x), then apply f(x) = tan(x) to both sides:
$$tan(y) = x$$
Now taking the derivative of both sides gives:
$$sec^2(y) \frac{dy}{dx} = \frac{dx}{dx} = 1$$
Then we divide by sec^{2}(x) to get
$$\frac{dy}{dx} = \frac{1}{sec^2(y)}$$
Now using tan(y) = x, we can construct another triangle:
... and from that we find
$$\frac{d}{dx} tan^{1}(x) = \frac{1}{1 + x^2}$$
This derivative is calculated in much the same way. We'll skip the details for this one; you should try it on your own. The result is:
$$\frac{d}{dx} cos^{1}(x) = \frac{1}{\sqrt{1  x^2}}$$
You could use the same method to find derivatives of the inverse cosecant, secant and cotangent functions, too. Try it!
$$f(x) = sin(x) \:\: \& \:\: f^{1}(x) = sin^{1}(x) $$
We need the derivative of the function
$$f'(x) = cos(x)$$
Then it's just a matter of plugging the inverse in to cos(x):
$$f'(f^{1}(x)) = \frac{1}{cos(sin^{1}(x))}$$
Now it's a little difficult to convert this into the form we found in example 1, but if we plot the two, the result is in the graph on the right.
I've shifted the purple curve upward by 0.1 so you can see that the curves would perfectly overlap otherwise.
This graph is another version of the $f(x) = x^2$ , $f^{1}(x) = \sqrt{x}$ graph above. Point (2, 4) is shown on f(x), and its mirror image across y = x, (4, 2) is shown on f^{1}(x).
Let's say we want to know the derivative (slope) of the inverse function at x = 4, but we don't actually know the inverse function (I know we know it here, but pretend we don't). Turns out we don't really need to know f^{1}(x).
If (4, 2) is a point on f^{1}(x), then (2, 4) is the point on f(x) at which f(x) has the reciprocal slope.
The green lines are tangent to the functions at those points. They have reciprocal slope. That means that the slope of the inverse at x = 4 can be found by taking the derivative of f(x) at x = 2.
We saw above that if points (a, b) and (c, d) lie on f(x), then points (b, a) and (d, c) will lie on f^{1}(x). We can use those two points to calculate the slopes of the segments connecting those lines. On the function, that slope is
$$m_f = \frac{d  b}{c  a}$$
and on the inverse it is.
$$m_{f^{1}} = \frac{c  a}{d  b}$$
It's easy to recognize that these are reciprocals of each other:
$$m_f = \frac{1}{m_{f^{1}}}$$
locations a, b, c and d are completely arbitrary, so we could make the distance between our two points as close to zero as we want, as we would in taking a derivative, so the relationship between derivatives is:
$$\frac{d}{dx}f(x) = \frac{1}{\frac{d}{dx} f^{1}(x)}$$
1. 
Let $f(x) = 2x^5 + x^3 + 1$. Find $\frac{d}{dx}f^{1}(x)$ at x = 4.

2. 
Given that $f(x) = x^3 + 7x + 2$ and $f(1) = 10$, calculate the value of $f^{1}(10)$.

3. 
Given that $f(x) = 7x^3 + (ln(x))^3$ and $f(1) = 7$, calculate the value of $f^{1}(7)$.

4. 
Find the equation of the line tangent to the inverse of $f(x) = x^5 + 2x^3 + x  4$ at the point (4, 0).

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 $(f^{1})'(3)$. You will likely encounter problems like this on the AP calculus exam.

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