Timeline for A visualization for the quotient rule

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Jun 18 at 22:22	comment	added	Michael Hardy		@user52817 : But that is not in Dave Marain's answer.
Jun 16 at 17:29	comment	added	user52817		The equality tells us $\lim_{h\to0}\frac{(f/g)(x+h)-(f/g)(x)}{h}=\lim_{h\to0}\frac{(f(x+h)-f(x))g(x)-f(x)(g(x+h)-g(x))}{hg(x)^2}$, so the quotient is differentiable.
Jun 15 at 18:50	comment	added	Michael Hardy		@user52817 : And as I said, nothing in Dave Marain's answer implies that if $f$ and $g$ are differentiable, then so is $f/g.$ Rather, it shows that the expression given by the quotient rule is the derivative IF $f/g$ is differentiable.
Jun 14 at 22:39	comment	added	user52817		From the answer by Dave Marain: $f' = \left( \frac{f}{g} g \right)' = \left( \frac{f}{g} \right) g' + \left( \frac{f}{g} \right)' g \Rightarrow \frac{f' - \left( \frac{f}{g} \right) g'}{g} = \left( \frac{f}{g} \right)' \Rightarrow \left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2}$. The derivative of the quotient exists iff the algebraic expression involving limits $\frac{f'g - fg'}{g^2}$ exists and $g\ne0$. This algebraic combination of limits exists if $f$ and $g$ are differentiable.
Jun 14 at 6:35	comment	added	Tommi		This is true, but unfortunately does not really answer the question.
Jun 13 at 18:38	history	answered	Michael Hardy	CC BY-SA 4.0