product rule

The product rule states that if $f:ℝ\to ℝ$ and $g:ℝ\to ℝ$ are functions in one variable both differentiable at a point $x_0$, then the derivative of the product of the two functions, denoted $f\cdot g$, at $x_0$ is given by

$$\frac{\mathrm{d}}{\mathrm{d}x}\left(f\cdot g\right)(x_0)=f(x_0)g^{\prime }(x_0)+f^{\prime }(x_0)g(x_0).$$

Proof

See the proof of the product rule (http://planetmath.org/ProofOfProductRule).

0.1 Generalized Product Rule

More generally, for differentiable functions $f_1,f_2,\mathrm{\dots },f_n$ in one variable, all differentiable at $x_0$, we have

$D(f_1\mathrm{\cdots }{f}_n)(x_0)={\displaystyle \sum _{i=1}^n}\left(f_i(x_0)\mathrm{\cdots }{f}_{i-1}(x_0)\cdot D{f}_i(x_0)\cdot f_{i+1}(x_0)\mathrm{\cdots }{f}_n(x_0)\right).$

Also see Leibniz’ rule (http://planetmath.org/LeibnizRule).

Example

The derivative of $x\ln |x|$ can be found by application of this rule. Let $f(x)=x,g(x)=\ln |x|$, so that $f(x)g(x)=x\ln |x|$. Then $f^{\prime }(x)=1$ and $g^{\prime }(x)=\frac{1}{x}$. Therefore, by the product rule,

${\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}}(x\ln |x|)$ $=$ $f(x)g^{\prime }(x)+f^{\prime }(x)g(x)$

$=$ ${\displaystyle \frac{x}{x}}+1\cdot \ln |x|$

$=$ $\ln |x|+1$

Title	product rule
Canonical name	ProductRule
Date of creation	2013-03-22 12:27:57
Last modified on	2013-03-22 12:27:57
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	12
Author	mathcam (2727)
Entry type	Theorem
Classification	msc 26A06
Related topic	Derivative
Related topic	ProofOfProductRule
Related topic	ProductRule
Related topic	PowerRule
Related topic	ProofOfPowerRule
Related topic	SumRule
Related topic	ZeroesOfDerivativeOfComplexPolynomial