1. **Problem:** State Cauchy’s Mean Value Theorem. 2. **Statement:** Cauchy’s Mean Value Theorem states that if functions $f$ and $g$ are continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, and if $g'(x) \neq 0$ for all $x$ in $(a,b)$, then there exists some $c \in (a,b)$ such that: $$\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$$ 3. **Explanation:** This theorem generalizes the Mean Value Theorem by relating the rates of change of two functions. 4. **Important conditions:** - $f$ and $g$ must be continuous on $[a,b]$. - $f$ and $g$ must be differentiable on $(a,b)$. - $g'(x) \neq 0$ for all $x$ in $(a,b)$. 5. **Summary:** There exists at least one point $c$ in $(a,b)$ where the ratio of the derivatives equals the ratio of the increments of the functions over $[a,b]$. Final answer: $$\boxed{\exists c \in (a,b) \text{ such that } \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}}$$