Find the derivative of the inverse function

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SUMMARY

The discussion centers on finding the derivative of an inverse function, specifically for a function defined as f(x) = x + (1/3)[f(x)]^3. The participant derived the inverse function as (1/3)x^3 - x and calculated its derivative to be x^2 - 1. However, another participant pointed out that when switching variables, one must consider the ranges of the variables involved, noting that the range of f(x) is [-1, ∞) rather than (-∞, ∞). This highlights the importance of understanding the implications of variable ranges in inverse function calculations.

PREREQUISITES
  • Understanding of inverse functions and their properties
  • Knowledge of derivatives and differentiation techniques
  • Familiarity with the concept of one-to-one functions
  • Basic grasp of function ranges and their significance
NEXT STEPS
  • Study the properties of one-to-one functions in calculus
  • Learn about the implications of variable ranges in inverse functions
  • Explore the chain rule and its application in finding derivatives of composite functions
  • Investigate the relationship between a function and its inverse, particularly in terms of graphical representation
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Students and professionals in mathematics, particularly those studying calculus, as well as educators teaching concepts related to inverse functions and derivatives.

ludi_srbin
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So I need to find the derivative of the inverse function. I know that f is one-to-one and is continous. Also I know that f'(x)=1+[f(x)]^2. I found the inverse writing my equation like f(x)=x+(1/3)[f(x)]^3 then I switch the variables and get that my inverse function=(1/3)x^3 - x. Then I just take the derivative and end up with x^2 - 1. Is this correct?
 
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ludi_srbin said:
So I need to find the derivative of the inverse function. I know that f is one-to-one and is continous. Also I know that f'(x)=1+[f(x)]^2. I found the inverse writing my equation like f(x)=x+(1/3)[f(x)]^3 then I switch the variables and get that my inverse function=(1/3)x^3 - x. Then I just take the derivative and end up with x^2 - 1. Is this correct?

:blushing: First, sorry for my poor english.
I don't think you are right.Because when you switch the variables in the equation you have to pay attention to the ranges of all variables.You can see in your equation f(x)=x^2-1,the range of x is(-∞,∞),but the range of f(x) is [-1,∞).So what you do is not only switch the variables but also write the range of x at the end of your equation.:-p
 
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Hmmm...But isn't the range of f(x) (-infinity, infinity)?
 

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