True or False: If $g \circ f$ is differentiable at $x_0$, then $f$ is differentiable at $x_0$, and $g$ is differentiable at $y=f(x_0)$.
I don't understand why if $(g\circ f)(x_0)$ is differentiable, then the following is not true: $f$ is differentiable at $x_0$ and $g$ is differentiable at $y_0=f(x_0)$.
Will the first statement not be true and why?
The chain rule says that if $f$ is differentiable at $x_0$, and $g$ is differentiable at $y_0 = f(x_0)$, then $g \circ f$ is differentiable at $x_0$ and $$(g \circ f)'(x_0) = g'(y_0) f'(x_0)$$ This question is about reversing some of the implications in that statement.
If you compose a misbehaving function with a constant function, you can cover up its misbehaving. Try $f(x) = |x|$ and $x_0 = 0$. You know $f$ is not differentiable at $x_0$.
Now let $g(x) = 0$ for all $x$. Then $g'(0) = 0$. But also, $(g \circ f)(x) = 0$ for all $x$. This means $(g \circ f)'(x_0) = 0$, as well.
So here we have a situation where the composition is differentiable, but one of the components is not.