Solution
= Solution
\b[No.] Take $f(x)=|x|$, $a=0$, and $g(y)=y^2$. Then $f$ is not <differentiable> at zero, but $g\circ f=x^2$ is <differentiable> there.
= Solution
\b[No.] Take $f(x)=|x|$, $a=0$, and $g(y)=y^2$. Then $f$ is not <differentiable> at zero, but $g\circ f=x^2$ is <differentiable> there.