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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?

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    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 $(f\circ g)'(x) = f'\bigl((g(x)\bigr)\cdot g'(x)$.2011-11-19

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The chain rule tells us more than just what the derivative is.

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) .

Thus, it also tells you that the derivative exists.

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    @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.)2011-11-19