Derivatives of inverse functions

[Zaqiqu]

Boy oh boy, I thought the Chain Rule was frustrating. At least there is a wealth of information to help the unwary Calculus student.
I've spent the last three hours working on one problem. I've read the section on "Derivatives of Inverse Functions" from five different textbooks. I've watched the few videos I can find that discuss the inverses of algebraic - as opposed to the ubiquitous trigonometric - functions. I'm no closer to finding a method than I was yesterday, when I spent another two hours on the same problem.

The textbook gives a "real number" a and a "function" f(x). It says to use the function f and the given real number a to find (f^-1)'(a).

The problem is that the solution manual, the examples, and everything else I can find is skipping the ONLY step that is confusing me.

Take the problem
f(x) = x³ - 1, a=26
After much trial and error, I plug in
26 = x³-1
27 = x³
3 = x

and so

1/f'(x) = 1/(3(3)²) = 1/27

My problem is, for the more complicated questions that come after this (and even the question above I probably wouldn't have solved on my own, thanks to the ridiculously simplistic example (that's singular) given in the book), finding x when a=n is a hassle. Take, for instance,

f(x) = 1/27 (x⁵ + 2x³), a=-11

How do I find the output of f(x) to get the input of (f^-1)'(x)?

I'm totally lost here.

 

[stapel]

The textbook gives a "real number" a and a "function" f(x). It says to use the function f and the given real number a to find (f^-1)'(a).

The problem is that the solution manual, the examples, and everything else I can find is skipping the ONLY step that is confusing me.

Take the problem
f(x) = x³ - 1, a=26
After much trial and error, I plug in
26 = x³-1
27 = x³
3 = x

and so

1/f'(x) = 1/(3(3)²) = 1/27

My problem is, for the more complicated questions that come after this (and even the question above I probably wouldn't have solved on my own, thanks to the ridiculously simplistic example (that's singular) given in the book), finding x when a=n is a hassle.

Are you saying that you're having trouble finding the inverse functions?

Take, for instance,

f(x) = 1/27 (x⁵ + 2x³), a=-11

How do I find the output of f(x) to get the input of (f^-1)'(x)?

So you're needing to find the derivative of the inverse function at a = -11? And you're having trouble finding the value of x such that f(x) = a, so that f^-1(a) = x?

All you need to do is to use what you learned back in algebra about solving polynomial equations. (here) You have the equation:

. . . . .$\displaystyle -11\, =\, \dfrac{1}{27}\, (x^5\, +\, 2x^3)$

Follow the usual procedure: multiply through to clear the fraction, move everything over to one side of the "equals" sign, and start with the Rational Roots Test. Then test potential zeroes (for instance, with synthetic division) until you've found what you need. (Doing a quick graph on your graphing calculator tells you to start your testing with x = -3.)

Once you remove the first factor (which gives you the first solution), you can continue trying to find zeroes, or else you can use what you know about first and second derivatives (as they relate to the shapes of graphs) to prove that this is the only zero. And this will give you your value for x.

 

[Zaqiqu]

Are you saying that you're having trouble finding the inverse functions?


So you're needing to find the derivative of the inverse function at a = -11? And you're having trouble finding the value of x such that f(x) = a, so that f^-1(a) = x?

All you need to do is to use what you learned back in algebra about solving polynomial equations. (here) You have the equation:

. . . . .$\displaystyle -11\, =\, \dfrac{1}{27}\, (x^5\, +\, 2x^3)$

Follow the usual procedure: multiply through to clear the fraction, move everything over to one side of the "equals" sign, and start with the Rational Roots Test. Then test potential zeroes (for instance, with synthetic division) until you've found what you need. (Doing a quick graph on your graphing calculator tells you to start your testing with x = -3.)

Once you remove the first factor (which gives you the first solution), you can continue trying to find zeroes, or else you can use what you know about first and second derivatives (as they relate to the shapes of graphs) to prove that this is the only zero. And this will give you your value for x.

Okay, I can do that. That was chapter 4 or 5, I think, of Stewart's Precalculus.

The trouble with textbooks is that most things seem to exist in a vacuum. You learn about Descartes' Rule of Signs, then it never comes up again; learn about angular motion, it never comes up again. It would be helpful if there was a book that really got down and explained some of these connections. Then again, that may be what a teacher is for.