{"id":11911,"date":"2018-04-02T10:30:43","date_gmt":"2018-04-02T17:30:43","guid":{"rendered":"https:\/\/magoosh.com\/hs\/?p=11911"},"modified":"2018-04-01T10:30:53","modified_gmt":"2018-04-01T17:30:53","slug":"ap-calculus-review-derivatives-inverse-functions","status":"publish","type":"post","link":"https:\/\/magoosh.com\/hs\/ap\/ap-calculus-review-derivatives-inverse-functions\/","title":{"rendered":"AP Calculus Review: Derivatives of Inverse Functions"},"content":{"rendered":"

One of the trickiest topics on the AP Calculus AB\/BC exam is the concept of inverse functions<\/strong> and their derivatives. In this review article, we’ll see how a powerful theorem can be used to find the derivatives of inverse functions. Then we’ll talk about the more common inverses and their derivatives.<\/p>\n

What are Inverse Functions?<\/h2>\n

Basically, an inverse function<\/strong> is a function that “reverses” what the original function did.<\/p>\n

For example, consider f<\/em>(x<\/em>) = 3x – 6. What does f<\/em> do to its input x<\/em>? Using correct order of operations, f<\/em> has the following effect:<\/p>\n

    \n
  1. First multiply by 3.<\/li>\n
  2. Then subtract 6 from the result.<\/li>\n<\/ol>\n

    So, in order to reverse what f<\/em> does, we have to follow the steps backwards. This is very much like reversing your driving directions to return home from an unfamiliar place.<\/p>\n

      \n
    1. First add 6 (to undo the subtraction).<\/li>\n
    2. Then divide by 3 (to undo the multiplication).<\/li>\n<\/ol>\n

      Therefore, the inverse function, which we’ll call g<\/em>(x<\/em>) for right now, has the formula,<\/p>\n

      g<\/em>(x<\/em>) = (x<\/em> + 6)\/3<\/p>\n

      The notation for the inverse function of f<\/em> is f<\/em> -1<\/sup>. So we could write:<\/p>\n

      f<\/em> -1<\/sup>(x<\/em>) = (x<\/em> + 6)\/3<\/p>\n

      Our purpose here is not to be able to solve to find inverse functions in all cases. In fact, the main theorem for finding their derivatives does not require solving for f<\/em> -1<\/sup>(x<\/em>) explicitly.<\/p>\n

      Finding the Derivative of an Inverse Function<\/h2>\n

      Computing the derivative of an inverse function is not too much more difficult than computing derivatives in general. <\/p>\n

      First, here’s a quick review of the basic derivative rules: Calculus Review: Derivative Rules<\/a>.<\/p>\n

      The Main Theorem for Inverses<\/h3>\n

      Suppose that f<\/em> is a function that has a well-defined inverse f<\/em> -1<\/sup>, and suppose that (a<\/em>, b<\/em>) is a point on the graph of y<\/em> = f<\/em>(x<\/em>). Then<\/p>\n

      \"Derivatives<\/p>\n

      However, you might see a different version of this rule. Another way to say that (a<\/em>, b<\/em>) is a point on the graph of y<\/em> = f<\/em>(x<\/em>) is to say that b<\/em> = f<\/em>(a<\/em>). Moreover, by properties of the inverse, then we can say that a<\/em> = f<\/em> -1<\/sup>(b<\/em>).<\/p>\n

      Finally, replacing b<\/em> by x<\/em>, we discover the second version of the Main Theorem:<\/p>\n

      \"Derivative<\/p>\n

      The two versions are useful in different contexts, which we shall see in the examples.<\/p>\n

      It’s also good to know that the condition of “having a well-defined inverse” is satisfied whenever the function is one-to-one<\/a><\/strong>.<\/p>\n

      Example 1<\/h3>\n

      Suppose that g<\/em>(x<\/em>) is the inverse function for f<\/em>(x<\/em>) = 3x<\/em>5<\/sup> + 6x<\/em>3<\/sup> + 4. Find the value of g<\/em> '(13).<\/p>\n

      Solution<\/h4>\n

      It won’t do us any good to try to solve for the inverse function algebraically. This polynomial is just too complicated for that. Instead, we must rely on the Main Theorem.<\/p>\n

      First, find the derivative of the original function.<\/p>\n

      f<\/em> '(x<\/em>) = 15x<\/em>4<\/sup> + 18x<\/em>2<\/sup><\/p>\n

      The other piece of the puzzle is the value to plug in. Identify b<\/em> = 13 from the problem statement. But don’t plug 13 into anything! Instead, the formula requires a value a<\/em> such that f<\/em>(a<\/em>) = b<\/em>; that is, f<\/em>(a<\/em>) = 13. <\/p>\n

      So it looks like we might have to solve:<\/p>\n

      3x<\/em>5<\/sup> + 6x<\/em>3<\/sup> + 4 = 13<\/p>\n

      However, that polynomial is still way to difficult to solve algebraically. Fortunately, it doesn’t take long to guess and check a value for x<\/em> that would make this equation true. <\/p>\n

      Just look at the coefficients: they add up to 13… Thus, if we just plugged in x<\/em> = 1, then we’d get the correct result.<\/p>\n

      3(1)5<\/sup> + 6(1)3<\/sup> + 4 = 3 + 6 + 4 = 13 <\/p>\n

      This means we should use a<\/em> = 1 in the formula for the derivative.<\/p>\n

      Finally, put it all together using the formula from the Main Theorem:<\/p>\n

      g<\/em> '(b<\/em>) = 1\/f<\/em> '(a<\/em>)<\/p>\n

      g<\/em> '(13) = 1\/f<\/em> '(1) = 1\/(15(1)4<\/sup> + 18(1)2<\/sup>) = 1\/33.<\/p>\n

      Common Inverses and Their Derivatives<\/h2>\n

      Some functions are so important that their inverses get special names. We’ll take a look at two such cases:<\/p>\n