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  "body": "\n## Proof\n\nWe write $f/g = f \\cdot (1/g)$ near $a$ and use the [[product rule|0V9A43H7WV]] and the [[reciprocal rule|0QD4E5JYPT]].\n\n1. Since $g$ is differentiable at $a$, it is continuous at $a$ by [[differentiable functions are continuous|07ZMX4ANJS]]. By the [[epsilon-delta form of continuity|0W8YHHGCJG]] with $\\varepsilon = \\abs{g(a)}$, there is $\\delta_0 \u003e 0$ with $\\abs{g(x) - g(a)} \u003c \\abs{g(a)}$ for $x \\in I$ with $\\abs{x - a} \u003c \\delta_0$.\n2. For such $x$, the [[reverse triangle inequality|0C9S13WKMF]] gives $\\abs{g(x)} \\ge \\abs{g(a)} - \\abs{g(x) - g(a)} \u003e 0$.\n3. Let $I_0 = I \\cap (a - \\delta_0, a + \\delta_0)$. It is an open interval containing $a$, and $g$ has no zero on $I_0$. So $f/g$ is defined on $I_0$.\n4. Claim: if $u \\colon I \\to \\R$ is differentiable at $a$, then its restriction to $I_0$ is differentiable at $a$ with the same [[derivative|0F1WR9C83K]].\n5. Proof of the claim. Let $H = \\set{h \\ne 0 : a + h \\in I}$ and $H_0 = \\set{h \\ne 0 : a + h \\in I_0}$. Then $H_0 \\subseteq H$, and $0$ is a [[limit point|07EGCNHAVW]] of $H_0$.\n6. Every sequence in $H_0$ tending to $0$ is a sequence in $H$. By the [[sequential characterization of limits|000MXHEY22]], the difference quotients of $u$ along it tend to $u'(a)$. These are also the difference quotients of the restriction. By the sequential characterization again, the claim follows.\n7. By step 4, $f$ and $g$ restricted to $I_0$ are differentiable at $a$. Since $g(a) \\ne 0$, the reciprocal rule on $I_0$ shows that $1/g$ is differentiable at $a$ with $(1/g)'(a) = -\\frac{g'(a)}{g(a)^2}$.\n8. On $I_0$ we have $\\frac{f}{g} = f \\cdot \\frac{1}{g}$. The product rule on $I_0$ shows that $f/g$ is differentiable at $a$, with\n$$\\left( \\frac{f}{g} \\right)'(a) = f'(a) \\cdot \\frac{1}{g(a)} + f(a) \\cdot \\left( -\\frac{g'(a)}{g(a)^2} \\right).$$\n9. Over the common denominator $g(a)^2$, this equals $\\frac{f'(a) g(a) - f(a) g'(a)}{g(a)^2}$.\n\n## Idea\n\nDivision by $g$ is multiplication by $1/g$. The product rule handles the multiplication, and the reciprocal rule gives the derivative of $1/g$. The only extra care is to work on an interval around $a$ where $g$ has no zero.\n",
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