Difference between revisions of "Product rule for derivatives"
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(Created page with "The product rule for differentiation is the formula $$\dfrac{d}{dx}[f(x)g(x)] = f'(x)g(x)+f(x)g'(x).$$") |
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| − | + | ==Theorem== | |
| − | $$\dfrac{d}{ | + | Let $f$ and $g$ be [[differentiable]] functions. Then, |
| + | $$\dfrac{\mathrm{d}}{\mathrm{d}x} \left[ f(x)g(x) \right] = f'(x)g(x) + f(x)g'(x),$$ | ||
| + | where $\dfrac{\mathrm{d}}{\mathrm{d}x}$ denotes the [[derivative|derivative operator]]. | ||
| + | |||
| + | ==Proof== | ||
| + | |||
| + | ==References== | ||
| + | * {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Sum rule for derivatives|next=Quotient rule for derivatives}}: $3.3.3$ | ||
Latest revision as of 17:21, 27 June 2016
Theorem
Let $f$ and $g$ be differentiable functions. Then, $$\dfrac{\mathrm{d}}{\mathrm{d}x} \left[ f(x)g(x) \right] = f'(x)g(x) + f(x)g'(x),$$ where $\dfrac{\mathrm{d}}{\mathrm{d}x}$ denotes the derivative operator.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $3.3.3$
