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Wardub
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 Derivative help   « on: Jul 21st, 2010, 10:04am » Quote Modify

(A^-1)' (x) = 1/(A'(A^-1(x))

That should read the derivative of A inverse.  I'm trying to follow a proof and can't understand how he got this.
It seems like it should involve the chain rule.  Can someone help me break it down step by step?

Thanks.
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0.999...
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 Re: Derivative help   « Reply #1 on: Jul 21st, 2010, 10:18am » Quote Modify

Assuming the existence of an inverse of A, we have the equation,
A(A-1(x)) = x .
Now, we implicitly differentiate w.r.t. x and indeed the chain rule implies that
(A-1)'(x)*A'(A-1(x)) = 1 .
Hence the result.

Visually, since the graph of the inverse function A-1 is a reflection of the initial function A across y = x, if at (a,b) A has slope dy/dx then at (b,a) A-1 has slope dx/dy.
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