# 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:

$$\phi (y)=\{\begin{array}{cc}\frac{f(y)-f({y}_{0})}{y-{y}_{0}}\hfill & \text{if}y\ne {y}_{0}\hfill \\ {f}^{\prime}({y}_{0})\hfill & \text{if}y={y}_{0}\hfill \end{array}$$ |

Since $f$ is differentiable in ${y}_{0}$, $\phi $ is continuous^{}.
We observe that, for $x\ne {x}_{0}$,

$$\frac{f(g(x))-f(g({x}_{0}))}{x-{x}_{0}}=\phi (g(x))\frac{g(x)-g({x}_{0})}{x-{x}_{0}},$$ |

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

Since $g$ is continuous in ${x}_{0}$, and $\phi $ is continuous in ${y}_{0}$,

$$\underset{x\to {x}_{0}}{lim}\phi (g(x))=\phi (g({x}_{0}))={f}^{\prime}(g({x}_{0})),$$ |

hence

${(f\circ g)}^{\prime}({x}_{0})$ | $=$ | $\underset{x\to {x}_{0}}{lim}{\displaystyle \frac{f(g(x))-f(g({x}_{0}))}{x-{x}_{0}}}$ | ||

$=$ | $\underset{x\to {x}_{0}}{lim}\phi (g(x)){\displaystyle \frac{g(x)-g({x}_{0})}{x-{x}_{0}}}$ | |||

$=$ | ${f}^{\prime}(g({x}_{0})){g}^{\prime}({x}_{0}).$ |

Title | proof of chain rule |
---|---|

Canonical name | ProofOfChainRule |

Date of creation | 2013-03-22 12:41:48 |

Last modified on | 2013-03-22 12:41:48 |

Owner | n3o (216) |

Last modified by | n3o (216) |

Numerical id | 6 |

Author | n3o (216) |

Entry type | Proof |

Classification | msc 26A06 |