My text asks one to prove that if $f$ is differentiable at $a$ then $\left|f\right|$ is differentiable at $a$, $a\ne0$. I understand how to do this using the definition of the derivative, which the answer key confirms; but the key then goes on to suggest that "it is also possible to use the chain rule", starting with $\left|{f}\right|=\sqrt{f^{2}}$ so that \left|{f}\right|\left(x\right)'=\frac{1}{2\sqrt{f\left(x\right)^{2}}}\cdot 2 f\left(x\right)f'\left(x\right) = f'\left(x\right) \cdot \frac{f\left(x\right)}{\left|f\left(x\right)\right|}\text{ .} But doesn't this simply show what the derivative will be if $\left|f\right|$ is differentiable? To use this method to prove that it is, wouldn't I first have to prove (in addition to already knowing that $f$ is differentiable at $a$) that $g\left(x\right)=x^{2}$ is differentiable at $f\left(a\right)$, that $h\left(x\right)=\sqrt{x}$ is differentiable at $g\left(f\left(a\right)\right)$?
In suggesting using the chain rule in this way, hasn't the answer key confused showing what something is, if it exists, with proving that something exists in the first place?