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        riddles >> putnam exam (pure math) >> Derivative help
        
(Message started by: Wardub on Jul 21^st, 2010, 10:04am)

Title: Derivative help
Post by Wardub on Jul 21^st, 2010, 10:04am

(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.

Title: Re: Derivative help
Post by 0.999... on Jul 21^st, 2010, 10:18am

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.