# proof of product rule

We begin with two differentiable functions $f(x)$ and $g(x)$ and show that their product is differentiable^{}, and that the derivative of the product has the desired form.

By simply calculating, we have for all values of $x$ in the domain of $f$ and $g$ that

$\frac{\mathrm{d}}{\mathrm{d}x}}\left[f(x)g(x)\right]$ | $=$ | $\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)g(x+h)-f(x)g(x)}{h}}$ | ||

$=$ | $\underset{h\to 0}{lim}{\displaystyle \frac{f(x+h)g(x+h)+f(x+h)g(x)-f(x+h)g(x)-f(x)g(x)}{h}}$ | |||

$=$ | $\underset{h\to 0}{lim}\left[f(x+h){\displaystyle \frac{g(x+h)-g(x)}{h}}+g(x){\displaystyle \frac{f(x+h)-f(x)}{h}}\right]$ | |||

$=$ | $\underset{h\to 0}{lim}\left[f(x+h){\displaystyle \frac{g(x+h)-g(x)}{h}}\right]+\underset{h\to 0}{lim}\left[g(x){\displaystyle \frac{f(x+h)-f(x)}{h}}\right]$ | |||

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

The key argument here is the next to last line, where we have used the fact that both $f$ and $g$ are differentiable, hence the limit can be distributed across the sum to give the desired equality.

Title | proof of product rule |
---|---|

Canonical name | ProofOfProductRule |

Date of creation | 2013-03-22 12:28:00 |

Last modified on | 2013-03-22 12:28:00 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 6 |

Author | mathcam (2727) |

Entry type | Proof |

Classification | msc 26A24 |

Related topic | Derivative |

Related topic | ProductRule |