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  "body": "\n## Proof\n\nWe first find an open interval around $a$ where $1/g$ is defined. Then we use the [[sequential characterization of limits|000MXHEY22]].\n\n1. Since $g$ is differentiable at $a$, it is continuous at $a$ by [[differentiable functions are continuous|07ZMX4ANJS]]. Put $\\eta = \\abs{g(a)} \u003e 0$.\n2. By the [[epsilon-delta form of continuity|0W8YHHGCJG]], there is $\\delta_0 \u003e 0$ with $\\abs{g(x) - g(a)} \u003c \\eta$ for $x \\in I$ with $\\abs{x - a} \u003c \\delta_0$.\n3. For such $x$, the [[reverse triangle inequality|0C9S13WKMF]] gives $\\abs{g(x)} \\ge \\abs{g(a)} - \\abs{g(x) - g(a)} \u003e 0$. So $g(x) \\ne 0$.\n4. Let $I_0 = I \\cap (a - \\delta_0, a + \\delta_0)$. It is an open interval containing $a$, and $g$ has no zero on it. So $1/g$ is defined on $I_0$, and we view it as a function $I_0 \\to \\R$.\n5. Let $H = \\set{h \\ne 0 : a + h \\in I_0}$. Since $I_0$ is an open interval containing $a$, $0$ is a [[limit point|07EGCNHAVW]] of $H$.\n6. For $h \\in H$, write $q_g(h) = \\frac{g(a + h) - g(a)}{h}$. Over a common denominator,\n$$\\frac{1}{h}\\left( \\frac{1}{g(a + h)} - \\frac{1}{g(a)} \\right) = \\frac{- q_g(h)}{g(a + h) \\, g(a)}.$$\n7. Let $(h_n)$ be a sequence in $H$ with $h_n \\to 0$. Each $h_n$ is a nonzero number with $a + h_n \\in I$. By the definition of the [[derivative|0F1WR9C83K]] and the sequential characterization, $q_g(h_n) \\to g'(a)$.\n8. Claim: $g(a + h_n) \\to g(a)$. Let $\\varepsilon \u003e 0$, and take $\\delta \u003e 0$ from the epsilon-delta form of continuity of $g$ at $a$. By the [[epsilon-N form|09G6KHS6GY]], $\\abs{h_n} \u003c \\delta$ for all large $n$. Then $\\abs{g(a + h_n) - g(a)} \u003c \\varepsilon$.\n9. The constant sequences $g(a)$ and $-1$ converge to $g(a)$ and $-1$. By the [[limit of a product of sequences|0734RSVF2V]], $g(a + h_n) \\, g(a) \\to g(a)^2$ and $-q_g(h_n) \\to -g'(a)$.\n10. We have $g(a)^2 \\ne 0$, and $g(a + h_n) g(a) \\ne 0$ for all $n$ by step 4. By the [[limit of a quotient of sequences|03S6GWWKEZ]], the right side of step 6 at $h = h_n$ tends to $-\\frac{g'(a)}{g(a)^2}$.\n11. This holds for every such sequence. By the sequential characterization, the difference quotient of $1/g$ at $a$ tends to $-\\frac{g'(a)}{g(a)^2}$ as $h \\to 0$.\n12. So $1/g$ is [[differentiable|0A61B60B5R]] at $a$, and $\\left( \\frac{1}{g} \\right)'(a) = -\\frac{g'(a)}{g(a)^2}$.\n\n## Idea\n\nSubtracting two fractions puts $g(a + h) - g(a)$ in the numerator and $g(a + h) g(a)$ in the denominator. The first gives the derivative of $g$ after dividing by $h$. The second tends to $g(a)^2$, which is not zero, because $g$ is continuous at $a$.\n",
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