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True of False: If $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, then $(g\circ f)''(a)=g'(f(a))f''(a)+g''(f(a))(f'(a))^2$.

I wasn't sure if my interpretation of this problem was correct. Is this the Chain Rule twice differentiable at $a$? If so, then it is indeed true but I'm not sure how to get this proof started. Any suggestions would be greatly appreciated.

My book provides a proof for the regular chain rule but attempting to follow along that and adjust the details to fit this question wasn't working out.

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    $\begingroup$ $f$ may not be twice differentiable, jus pick $f(x)=x|x|$ and you can show the statement is not correct. $\endgroup$
    – John
    Oct 24, 2014 at 5:23

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This is true as long as both $f$ and $g$ are atleast twice differentiable, use the $\color{red}{\mbox{Chain Rule}}$ twice and the $\color{blue}{\mbox{Product rule}}$.

$$ ((f \circ g)(a))^{\prime\prime} = \left( g(f(a)) \right)^{\prime\prime} = \left( \color{red}{ g^{\prime}(f(a))f^{\prime}(a) } \right)^{\prime} $$

Then apply the Product Rule on $$\color{maroon}{ g^{\prime}(f(a))}\color{OrangeRed}{f^{\prime}(a)} $$ and then the Chain Rule on $$ \color{red}{ g^{\prime}(f(a))}f^{\prime}(a) \;\mbox{ then again on } \; \color{maroon}{ g^{\prime}(f(a))} $$.

$$ \big(\color{maroon}{ g^{\prime}(f(a))}\color{OrangeRed}{f^{\prime}(a)} \big)^{\prime} = \color{maroon}{ g^{\prime\prime}(f(a))f^{\prime}}\color{OrangeRed}{f^{\prime}(a)} \color{blue}{ + } \color{maroon}{ g^{\prime}(f(a))}\color{OrangeRed}{f^{\prime\prime}} = g^{\prime\prime}(f(a))(f^{\prime}(a))^2 + g^{\prime}(f(a))f^{\prime\prime} $$

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