## You are here

Homeproof of product rule

## Primary tabs

# 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

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

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

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

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

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

Related:

Derivative, ProductRule

Type of Math Object:

Proof

Major Section:

Reference

Parent:

Groups audience:

## Mathematics Subject Classification

26A24*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff

## Recent Activity

Sep 14

new problem: Geometry by parag

Aug 24

new question: Scheduling Algorithm by ncovella

new question: Scheduling Algorithm by ncovella

new problem: Geometry by parag

Aug 24

new question: Scheduling Algorithm by ncovella

new question: Scheduling Algorithm by ncovella