The first good news is that even though there is no general way to compute the value of the inverse to a function at a given argument, there is a simple formula for the derivative of the inverse of f in terms of the derivative of f itself.
In fact, the derivative of f -1 is the reciprocal of the
derivative of f, with argument and value reversed.
This is more or less obvious geometrically. You get the graph of the inverse of f from that of the function f by switching x and y axes. The derivative of anything is the change of the function (y) divided by the change of the argument, x. Switching value and argument means that the switched derivative becomes the change of the original argument, x, divided by the change of the original value, y. This is the reciprocal of the original derivative, but at argument given by the value (the original x) of the inverse function. Draw a picture and look and you will agree.
Let's prove this using algebra. All we have to do is to apply the chain rule to the defining property of f -1, which is f -1(f(x)) = x and f(f -1(x)) = x.
The derivative of the right hand side is 1 here in both equations. The derivative of the left hand side in the second case is, by the chain rule, the product of the derivative that we want of f -1 and the derivative of f evaluated at the value of f -1(x) This is exactly the statement that the derivative of f -1 is the reciprocal of the derivative of f evaluated at the value rather than the argument of f-1.
The argument seems simple enough but it is confusing. Can you use this rule to actually find derivatives of inverses without going nuts?
Let us see what this means for the exponential function and its inverse, ln(x). The derivative of the exponent function is itself, exp(x). Then the derivative of the logarithm is the reciprocal of the exponent, evaluated at ln(x) or . The latter identity follows from the definition of inverse which tells us that exp(ln(x) = x.
Similarly, for the sine function, since its derivative at argument x is cos(x), the derivative of arcsin(x) is . You could leave it at that, but we generally reduce it to something slightly less ugly. A spreadsheet is as happy with this as with the result we end up with in the next paragraph. By the way, my spreadsheet knows the arcccosine function of argument A6 as =acos(A6).
By the Pythagorean theorem of plane geometry we have . If we set y = arcsin(x), we get from this statement: . Since sin(arcsin(x)) is x, we finally get, that the derivative of arcsin(x) is .
Similarly the derivative of . The derivative of which is the inverse to is then the reciprocal of this expression or where y is the value of the inverse function, which is . The derivative of is then , or, if you prefer, .
Notice that this formula is exactly the same as the power formula for integer powers. In fact for any rational power, a, positive or negative, we have
(xa)' = axa-1
There is another piece of good news about inverse functions. Even though there is no obvious way to compute a particular value of one, at a particular argument, there is any easy way to compute the value of f-1(x) with a spreadsheet that you can actually perform in about a minute, once you know how, assuming you know how to compute f. We will see it soon.
8.3 Use the fact that is true, find . (you can use the multiple occurrence rule)
8.4 The tangent of an angle z, denoted as tan z, is the ratio given by the sine divided by the cosine: . What is the derivative of tan z? From it find the derivative of atan z (called the arctangent of z) which is the inverse function to tan z, (when the domain of tan z has been restricted to be from .)
8.5 Plot tan x and atan x using the applet that takes inverses.
I am getting tired of this stuff.
Well, we really are done with what is traditionally taught about how to differentiate functions. The multiple occurrence rule, the chain rule and the inverse rule tell us how to differentiate anything we can construct, starting from the three functions x, exp x, and sin x, whose derivatives we know. It is easy to make mistakes when you find derivatives, so it is wise to have a way to check your answers.
An easy way is to compare them to the results of differentiating numerically, which we now describe.