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September 2nd, 2014, 04:43 AM   #1
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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.
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September 2nd, 2014, 04:55 AM   #2
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By the chain rule:$$G(x) = \frac{1}{g(x)} \implies G^\prime(x) = -\frac{1}{g^2(x)} g^\prime(x)$$
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September 2nd, 2014, 04:57 AM   #3
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$\displaystyle G(x)=\frac{1}{g(x)} \Rightarrow G'(x)=-\frac{g'(x)}{g(x)^2}$

$\displaystyle \left(f^{-1}\right)'(f(3))=\frac{1}{f'(3)}\Rightarrow g'(5)=\frac{1}{\frac{1}{9}}=9$

$\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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