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.

  • 2
    $\begingroup$ $f$ 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

1 Answer 1


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 $$ \color{red}{ g^{\prime}(f(a))}f^{\prime}(a) \;\mbox{ then again on } \; \color{maroon}{ g^{\prime}(f(a))} $$.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.