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Let $E$ be an open subset of $\mathbb{R}^n$ and $f,g:E\to \mathbb{R}^1$ be differentiable functions with $g\neq 0$ on $E$. How to prove that $\dfrac{f}{g}$ is also differentiable?

I can't prove this rigorously. Can anyone show the full proof?

I would be very thankful!

Raheem Najib
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1 Answers1

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Let $\phi(x) = {1 \over x}$. Then for $x \neq 0$ we have ${\phi(x+h)-\phi(x) \over h} = - {1 \over x^2+xh}$, and we see that $\lim_{h \to 0} {\phi(x+h)-\phi(x) \over h} = - {1 \over x^2}$, so $\phi$ is differentiable.

Hence $(\phi \circ g)(x) = { 1\over g(x)}$ is differentiable by the chain rule, and so ${f(x) \over g(x)} = f(x) \cdot (\phi \circ g)(x) $ is differentiable by the product rule.

For the product rule, if $f,g$ are differentiable at $x$, then \begin{eqnarray} { f(x+h)g(x+h)-f(x)g(x) \over h} &=& { f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x) \over h} \\ &=& f(x+h) { g(x+h)-g(x) \over h} + { f(x+h)-f(x) \over h} g(x) \\ \end{eqnarray}

Hence, since $f$ is continuous at $x$ we have $\lim_{h \to 0} { f(x+h)g(x+h)-f(x)g(x) \over h} = f(x)g'(x)+f'(x)g(x)$.

copper.hat
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  • That's nice but how to prove product rule? Can you show it in your above post if it's not difficult? – Raheem Najib Feb 18 '16 at 05:02
  • @RaheemNajib: I added a proof of the product rule. – copper.hat Feb 18 '16 at 05:45
  • @copper.hat Thank you for this proof, it helped my understanding a lot, but I think it could be improved if it was more clear where the "multivariable" part went. For example, I don't think what is written for the product rule makes sense in this context, because the domains of f and g are open subsets of $\mathbb{R}^n$, not $\mathbb{R}$, so for example the $h$ in the denominator should be $|h|$, but then a different definition of derivative needs to be used. One easy solution would be to just delete the proof of the product rule. – CJD Apr 16 '19 at 15:39
  • @CJD: You are correct, I will correct this when I have time. – copper.hat Apr 16 '19 at 15:55