# General Theorem # Others > [!NOTE] Definition > Let $f: U\subset \mathbb{R}^{n}\to \mathbb{R}$ be a [[Real-Valued Function on Real n-Space (Multivariable Function)|multivariable function]]. Let $\underline{x}:I \subset \mathbb{R}\to U$ be a [[Vector-Valued Function of Real variable|vector-valued function of single variable]] that is [[Real Differentiablity|differentiable]] at some $c\in I.$ If the [[Partial Derivatives (of Real-Valued Function on Real n-Space)|partial derivatives]] of $f$ exist at $\underline{x}(c),$ then the [[Derivative of Real Function|derivative]] of $f\circ x: \mathbb{R}\to \mathbb{R}$ with respect to $t$ at $c$ exists and is given in by $\frac{df}{dt}(c)=\sum \frac{ \partial f }{ \partial x } \frac{dx_{i}}{dt} (c)$ **Proof**: ... > [!NOTE] Theorem (Chain Rule) > Suppose $I$ and $J$ are [[Real Interval|open intervals]], $f:I\to \mathbb{R}$ and $g:J\to I$ such that $g$ is [[Real Differentiablity|differentiable]] at $c$ and $f$ is differentiable at $g(c)$. Then the composition $f\circ g$ is differentiable at $c$ and $(f \circ g)'(c)=f'(g(c)) \cdot g'(c).$ **Proof**: Since $f$ is differentiable at $g(c),$ by [[Weirestrass-Caratheodory Criterion for Differentiability of Real Function|local linearisation]], for all $y\in I$, there exists a function $\varepsilon$ such that $f(y)-f(g(c))= f'(g(c))(y-g(c)) + \varepsilon(y)(y-g(c))$where $\varepsilon(g(c))=0$ and $\varepsilon(x) \to 0$ as $x\to(g(c))$. Hence $f(g(x))-f(g(c))= f'(g(c))(g(x)-g(c)) + \varepsilon(g(x))(g(x)-g(c)).$Consequently if $x \neq c$, $\frac{f(g(x))-g(c)}{x-c} = f'(g(c)) \frac{g(x)-g(c)}{x-c} + \varepsilon (g(x)) \frac{g(x)-g(c)}{x-c}$As $x\to c$, $\frac{g(x)-g(c)}{x-c} \to g'(c)$while $\varepsilon \circ g$ is [[Continuous Real Function|continuous]] at $c$ so $\varepsilon(g(x))\to\varepsilon(g(c))=0$. Hence $\frac{f(g(x))-f(g(c))}{x-c} \to f'(g(c)) \cdot g'(c)$