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  "body": "\n## Proof\n\nWe apply Rolle's theorem to a suitable linear combination of $f$ and $g$.\n\n1. Let $A = f(b) - f(a)$ and $B = g(b) - g(a)$. These are real constants.\n2. Define $h \\colon [a, b] \\to \\R$ by $h(x) = A \\, g(x) - B \\, f(x)$.\n3. A linear combination of [[continuous|0P27W9RW0G]] functions is continuous, so $h$ is continuous on $[a, b]$.\n4. By the sum and [[constant multiple rule|0Z9JB71BS0]], $h$ is differentiable on $(a, b)$ with $h'(x) = A \\, g'(x) - B \\, f'(x)$.\n5. Expanding, $h(a) = f(b) g(a) - f(a) g(a) - g(b) f(a) + g(a) f(a) = f(b) g(a) - f(a) g(b)$.\n6. Expanding, $h(b) = f(b) g(b) - f(a) g(b) - g(b) f(b) + g(a) f(b) = f(b) g(a) - f(a) g(b)$.\n7. So $h(a) = h(b)$. By [[Rolle's theorem|0194DY35BG]], there is $c \\in (a, b)$ with $h'(c) = 0$.\n8. By step 4, $A \\, g'(c) = B \\, f'(c)$. That is, $(f(b) - f(a)) \\, g'(c) = (g(b) - g(a)) \\, f'(c)$.\n\n## Idea\n\nThe coefficients of $h$ make the endpoint values of $f$ and $g$ cancel, so $h(a) = h(b)$. Rolle's theorem then gives a point where $h' = 0$. At that point the claimed relation between $f'$ and $g'$ holds.\n",
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