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 September 2nd, 2014, 04:43 AM #1 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Inverse function of a differentiable function Suppose $\displaystyle g$is the inverse function of a differentiable function $\displaystyle f$and $\displaystyle G(x) = \frac{1}{g(x)}$ If $\displaystyle f(3)=5$ and $\displaystyle f'(3) = \frac{1}{9}$ find $\displaystyle G'(5)$ I know how to find $\displaystyle g'(5)$ Not sure how to find $\displaystyle G'(5)$ How does the reciprocal function tie into this. Not seeing how the solution has a negative number. Thanks for any help.
 September 2nd, 2014, 04:55 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,937 Thanks: 2265 Math Focus: Mainly analysis and algebra By the chain rule:$$G(x) = \frac{1}{g(x)} \implies G^\prime(x) = -\frac{1}{g^2(x)} g^\prime(x)$$ Thanks from ZardoZ
 September 2nd, 2014, 04:57 AM #3 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus $\displaystyle G(x)=\frac{1}{g(x)} \Rightarrow G'(x)=-\frac{g'(x)}{g(x)^2}$ $\displaystyle \left(f^{-1}\right)'(f(3))=\frac{1}{f'(3)}\Rightarrow g'(5)=\frac{1}{\frac{1}{9}}=9$ $\displaystyle \left(g\mathcal{o} f \right)(x)= x \Rightarrow \left(g\mathcal{o} f \right)(3)= 3\Rightarrow g\left(f(3)\right) =3 \Rightarrow g(5)=3$ $\displaystyle G'(5)=-\frac{9}{9}=-1$ Thanks from WWRtelescoping

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