Purpose of the derivative of the inverse function

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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.
 
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.