proof of chain rule

Let’s say that $g$ is differentiable in $x_{0}$ and $f$ is differentiable in $y_{0}=g(x_{0})$. We define:

 $\varphi(y)=\left\{\begin{array}[]{ll}\frac{f(y)-f(y_{0})}{y-y_{0}}&\textrm{if % y\neq y_{0}}\\ f^{\prime}(y_{0})&\textrm{if y=y_{0}}\end{array}\right.$

Since $f$ is differentiable in $y_{0}$, $\varphi$ is continuous. We observe that, for $x\neq x_{0}$,

 $\frac{f(g(x))-f(g(x_{0}))}{x-x_{0}}=\varphi(g(x))\frac{g(x)-g(x_{0})}{x-x_{0}},$

in fact, if $g(x)\neq g(x_{0})$, it follows at once from the definition of $\varphi$, while if $g(x)=g(x_{0})$, both members of the equation are 0.

Since $g$ is continuous in $x_{0}$, and $\varphi$ is continuous in $y_{0}$,

 $\lim_{x\to x_{0}}\varphi(g(x))=\varphi(g(x_{0}))=f^{\prime}(g(x_{0})),$

hence

 $\displaystyle(f\circ g)^{\prime}(x_{0})$ $\displaystyle=$ $\displaystyle\lim_{x\to x_{0}}\frac{f(g(x))-f(g(x_{0}))}{x-x_{0}}$ $\displaystyle=$ $\displaystyle\lim_{x\to x_{0}}\varphi(g(x))\frac{g(x)-g(x_{0})}{x-x_{0}}$ $\displaystyle=$ $\displaystyle f^{\prime}(g(x_{0}))g^{\prime}(x_{0}).$
Title proof of chain rule ProofOfChainRule 2013-03-22 12:41:48 2013-03-22 12:41:48 n3o (216) n3o (216) 6 n3o (216) Proof msc 26A06