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Letter representations:

• c =  any constant
• ƒ(x) and g(x) =  any function
• ^n = any power of a variable.  So x² is the same thing as x^2.  The program I’m writing with won’t let me superscript anything higher than 3, so x^n is the best I can do.
• ( ‘ or d/dx) = first derivative of a function (so u’, would mean the first derivative of function u, as would d/dx u)

Inverse function derivatives can be confusing when you first start out, but don’t worry!  When I took the AP AB Calculus test, they gave me all the parts I needed, just to see if I remembered the formula.

Remember, the inverse of a function is achieved when you switch x’s and y’s and solve for y.  With this in mind, let’s dive in!

Inverse Function Rule

Let g(x) be the inverse of ƒ(x).  So ƒ^-1(x) = g(x).

ƒ’ g(x) =      1    

               ƒ’(g(x))

Explanation:

Short Way (AP Test Tip):  You will usually be given information like: g(…) = … and ƒ’(…) = … in a word problem. on the AP AB Calculus exam (multiple choice) because it is a timed test to test your knowledge of calculus, not algebraic skills.  Doing this the long way takes way too much time for that kind of test.  Example AP test question:

Let g(x) be the inverse of ƒ(x).  If g(3) = 7.2 and ƒ’(7.2) = 4, what is ƒ’ g(3)?  (Note: there may be other “fluff” like the problem saying “The water tower has a drain rate of …. and the inverse of …blah blah blah.  Ignore that stuff.  Just key in on the word “inverse”, then look for the needed parts.

Plug these into the formula:

                    1        =        1        =     1 

               ƒ’(g(3))          ƒ’(7.2)          4

Easy, right?

Long Way:  Now suppose your teacher is not very nice (or just a good teacher that wants you to learn) and makes you do the entire process by hand, here is what you do.  (Note: my teacher did this, and the equation looked really nasty, but when solved it turned out to equal 2 or something like that.  Don’t be discouraged by a tough-looking problem – it may turn out to be really easy!)  Here’s an easy problem:

Let g(x) be the inverse of ƒ(x).  ƒ(x) = 2x + 1.  What is g’(5)?

First we need the inverse of ƒ(x).  So let’s get it

y = 2x + 1

Interchange x’s and y’s:

x = 2y + 1

Solve for y, then replace y with g(x) to remind yourself that this is the inverse:

g(x) = x − 1

             2

Now use the formula:

     1    

ƒ’(g(5))

The g(5) = 2, so plug that in

     1    

  ƒ’(2)

We need to find ƒ’ now.  ƒ’ = 2.  Wait a second.  There is no x to plug the g(5) = 2 into!  That’s not a problem.  Remember we just found the inverse of a power of 1 equation.  The derivative will not have an x.  So for all values of x, ƒ’ = 2.  Plug that in.

  1  

  2

½ is your answer.  For most of your problems that are given to you the long way, however, ƒ’ will have to be found and a value will have to be plugged in.  The last step there is really easy.  (I would have done it but I forgot that lines have no varying derivative when I made the problem up!  Sorry about that.)

Congratulations!  If you have been reading throughout all my articles on all 27 derivative rules, this was the last one!  My next series of articles will talk about the applications of derivatives, like implicit differentiation and related rates.  Until then, keep plugging and chugging away!