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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: 6,540 Thanks: 2146 Math Focus: Mainly analysis and algebra 
By the chain rule:$$G(x) = \frac{1}{g(x)} \implies G^\prime(x) = \frac{1}{g^2(x)} g^\prime(x)$$

September 2nd, 2014, 04:57 AM  #3 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 131 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 $ $\displaystyle G'(5)=\frac{9}{9}=1$ 

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differentiable, function, inverse 
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