[proofplan] We construct an auxiliary function $h$ that is a linear combination of $f$ and $g$ designed so that $h(a) = h(b) = 0$. Applying [Rolle's Theorem](/theorems/185) to $h$ produces a point $c \in (a,b)$ where $h'(c) = 0$. Differentiating $h$ and rearranging yields the Cauchy mean value equation. The denominator $g(b) - g(a)$ is nonzero because $g' \neq 0$ on $(a,b)$ forces $g(b) \neq g(a)$ by Rolle's Theorem (applied contrapositively). [/proofplan] [step:Verify that $g(b) \neq g(a)$] Suppose for contradiction that $g(b) = g(a)$. Then $g$ is [continuous](/page/Continuity) on $[a,b]$, [differentiable](/page/Derivative) on $(a,b)$, and $g(a) = g(b)$. By [Rolle's Theorem](/theorems/185), there exists $c \in (a,b)$ with $g'(c) = 0$, contradicting the hypothesis $g'(x) \neq 0$ for all $x \in (a,b)$. Therefore $g(b) \neq g(a)$, and the quotient $\frac{f(b) - f(a)}{g(b) - g(a)}$ is well-defined. [/step] [step:Construct the auxiliary function $h$ with equal endpoint values] Define \begin{align*} h: [a,b] &\to \mathbb{R} \\ x &\mapsto f(x) - f(a) - \frac{f(b) - f(a)}{g(b) - g(a)}\bigl(g(x) - g(a)\bigr). \end{align*} Since $f$ and $g$ are continuous on $[a,b]$ and differentiable on $(a,b)$, so is $h$. [guided] The construction mirrors the proof of the [Mean Value Theorem](/theorems/186), but instead of subtracting a linear function of $x$, we subtract a linear function of $g(x)$. This replaces the secant line in the $(x, f(x))$-plane with the secant line in the $(g(x), f(x))$-plane. Concretely, we define \begin{align*} h: [a,b] &\to \mathbb{R} \\ x &\mapsto f(x) - f(a) - \frac{f(b) - f(a)}{g(b) - g(a)}\bigl(g(x) - g(a)\bigr). \end{align*} The coefficient $\frac{f(b) - f(a)}{g(b) - g(a)}$ is chosen precisely so that $h(a) = h(b) = 0$: at $x = a$ the subtracted terms vanish ($g(a) - g(a) = 0$), and at $x = b$ the subtracted term equals $f(b) - f(a)$, cancelling the $f(b) - f(a)$ contribution. Since $f$ and $g$ are [continuous](/page/Continuity) on $[a,b]$ and [differentiable](/page/Derivative) on $(a,b)$, the function $h$ inherits both properties as a linear combination of continuous (resp. differentiable) functions. [/guided] [/step] [step:Apply Rolle's Theorem to $h$ and rearrange] Evaluating at the endpoints: \begin{align*} h(a) &= f(a) - f(a) - \frac{f(b) - f(a)}{g(b) - g(a)}\bigl(g(a) - g(a)\bigr) = 0, \\ h(b) &= f(b) - f(a) - \frac{f(b) - f(a)}{g(b) - g(a)}\bigl(g(b) - g(a)\bigr) = 0. \end{align*} By [Rolle's Theorem](/theorems/185), there exists $c \in (a,b)$ with $h'(c) = 0$. Differentiating: \begin{align*} h'(x) = f'(x) - \frac{f(b) - f(a)}{g(b) - g(a)} \cdot g'(x). \end{align*} Setting $h'(c) = 0$: \begin{align*} f'(c) = \frac{f(b) - f(a)}{g(b) - g(a)} \cdot g'(c). \end{align*} Since $g'(c) \neq 0$ by hypothesis, dividing both sides by $g'(c)$ gives \begin{align*} \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}. \end{align*} [/step]