# Proving differentiability by demonstrating what the derivative “is”

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?

-
Chain rule: If $g$ is differentiable at $x$ and if $f$ is differentiable at $g(x)$, then $f\circ g$ $is\ differentiable$ at $x$ and $(f\circ g)'(x) = f'\bigl((g(x)\bigr)\cdot g'(x)$. – David Mitra Nov 19 '11 at 15:22

The complete statement is that if $g$ is differentiable at $c$ (so $g'(c)$ exists) and $f$ is differentiable at $g(c)$, then $f ∘ g$ is differentiable at $c$, with its derivative given by $$(f ∘ g)'(c) = f'(g(c) ) \cdot g'(c) .$$
Yes "and $f$ is differentiable at $g(c)$" is what I was asking about (I think). – raxacoricofallapatorius Nov 19 '11 at 15:39
To the OP: You just need to assume $f(a)\ne0$ (not $a\ne0$ as in your post). Then, the chain rule applies. You'll have to use it twice. – David Mitra Nov 19 '11 at 15:57
@RagibZaman: Yes, so in order to use the chain rule I would need to assume (or prove) the additional conditions on (my, above) $g$, and $h$, right? (I understand that once I've shown I can use it, it will prove what I need to prove.) – raxacoricofallapatorius Nov 19 '11 at 16:26