Re: dif(f(x), x) = dif(x, f^-1(x))
From: Dave Langers (spam.messages_at_annoy.us)
Date: 11/22/04
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Date: Mon, 22 Nov 2004 21:08:02 +0100
>>Next, I'm not sure what you mean by f^-1(x). Do you mean the inverse
>>function of f such that
>> f(f^-1(x)) = x
>> f^-1(f(x)) = x
>>or do you mean
>> f^-1(x) = 1/(f(x))
> As I wrote
> f^-1(x) means the inverse of f(x), as if f(x) = y then x = f^-1(y)
Oops, stupid me.
The answer remains 'no'. However there is a nice way to calculate de
derivative of the inverse function, knowing the derivative of the
function itself. Suppose x = f(y) and y = f^-1(x)
d(f^-1(x))/dx = d (f^-1(f(y))) / d(f(y))
= dy/d(f(y))
= 1 / ( d(f(y))/dy )
So, for example, let's say that f(x) = tan(x) and you know that
d(tan(x))/dx = 1+tan^2(x), and you want to calculate the derivative of
it's inverse f(x) = arctan(x) which you don't know. Then from the above
d(arctan(x))/dx = 1 / ( d(tan(y))/dy ) = 1 / ( 1+tan^2(y) )
But since y = arctan(x) it reads
d(arctan(x))/dx = 1/(1+x^2)
Voila!
--
M.vr.gr.
Dave
("d-dot-langers-at-wxs-dot-nl")
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