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# 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

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## Mathematics Subject Classification

26A24*no label found*

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## Recent Activity

Jul 5

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias