# proof of quotient rule (using product rule)

Suppose $f$ and $g$ are differentiable functions defined on some interval of $\mathbb{R}$, and $g$ never vanishes. Let us prove that

$${\left(\frac{f}{g}\right)}^{\prime}=\frac{{f}^{\prime}g-f{g}^{\prime}}{{g}^{2}}.$$ |

Using the product rule^{} ${(fg)}^{\prime}={f}^{\prime}g+f{g}^{\prime}$, and ${({g}^{-1})}^{\prime}=-{g}^{-2}{g}^{\prime}$,
we have

${\left({\displaystyle \frac{f}{g}}\right)}^{\prime}$ | $=$ | ${(f{g}^{-1})}^{\prime}$ | ||

$=$ | ${f}^{\prime}{g}^{-1}+f{({g}^{-1})}^{\prime}$ | |||

$=$ | ${f}^{\prime}{g}^{-1}+f(-1){g}^{-2}{g}^{\prime}$ | |||

$=$ | $\frac{{f}^{\prime}}{g}}-{\displaystyle \frac{f{g}^{\prime}}{{g}^{2}}$ | |||

$=$ | $\frac{{f}^{\prime}g-f{g}^{\prime}}{{g}^{2}}}.$ |

Here ${g}^{-1}=1/g$ and ${g}^{-2}=1/{g}^{2}$.

Title | proof of quotient rule (using product rule) |
---|---|

Canonical name | ProofOfQuotientRuleusingProductRule |

Date of creation | 2013-03-22 15:00:45 |

Last modified on | 2013-03-22 15:00:45 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 5 |

Author | matte (1858) |

Entry type | Proof |

Classification | msc 26A06 |