Purpose of the derivative of the inverse function

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The derivative of the inverse function, expressed as g'(x) = 1/f'(g(x)), is crucial for understanding the relationship between a function and its inverse. A practical application is controlling engine speed with a dial, where small adjustments can be estimated using this derivative. The discussion also highlights the derivative of arccos(x), demonstrating its calculation through trigonometric identities. Participants seek additional examples of where this derivative is applicable beyond theoretical exercises. Understanding this concept is essential as it has widespread applications across various fields, including physics and engineering.
barryj
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Homework Statement


In calculus, I learn that the derivative of the inverse function is

g'(x) = 1/ f'(g(x))

Homework Equations


So..

The Attempt at a Solution



Can someone give me an example of where I need to know this, or is this just a math exercise. Is there a relatively simple physics example?
 
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Sure. Suppose you are controlling an engine speed with a dial. If you want to increase the speed by 3 rpm, how much should you increase the dial setting?

PS. Keep in mind that this gives a linear estimate at that setting. It maybe very different at another setting.
 
Don't go away for long. I will be back in4 hours.
 
This property of the derivative of the inverse function is just a simple observation of what a slope is when looked at from another direction. As such, it has applications everywhere.
 
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What is the derivative of ##\arccos x##? Good luck with using the definition.
 
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OK, finding the derivative of arccos(x)
given... f(x) = cos(X) and g(x) = cos^-1(x)

then g'(x) = 1/f'(g(x))

g'(x) = 1/-sin(g(x))

g'(x) = 1/-sin(cos^-1(x))

then after some trig substitutions we get this is equal to ##-1/\sqrt(1-u^2)##or something like this.
 
barryj said:
OK, finding the derivative of arccos(x)
given... f(x) = cos(X) and g(x) = cos^-1(x)

then g'(x) = 1/f'(g(x))

g'(x) = 1/-sin(g(x))

g'(x) = 1/-sin(cos^-1(x))

then after some trig substitutions we get this is equal to ##-1/\sqrt(1-u^2)##or something like this.

This is one of the numerous examples where it is useful.
 
Could someone give me a few more examples where finding the derivative of the inverse function is useful.
 

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