Product Rule

The product rule is useful for differentiating the product of functions.

It states that for any functions $u$ and $v$,

$\frac{ d}{dx} uv = u \frac{d}{dx} v + v \frac{d}{dx} u.$

Or, in Lagrange notation,

$(uv)' = u'v+v'u.$

We begin by recalling the definition of differentiation:

$\frac{d}{dx}f(x) = \lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}.$

Thus, we have

$\begin{aligned} \dfrac{d}{dx}(uv)&=\displaystyle\lim_{h\to 0}\dfrac{u(x+h)v(x+h)-u(x)v(x)}{h}\\ &=\displaystyle\lim_{h\to 0}\dfrac{u(x+h)v(x+h)-u(x+h)v(x)+u(x+h)v(x)-u(x)v(x)}{h}\\ &=\displaystyle\lim_{h\to 0}\left[u(x+h)\dfrac{v(x+h)-v(x)}{h}+v(x)\dfrac{u(x+h)-u(x)}{h}\right]\\ &=\displaystyle\lim_{h\to 0}u(x+h)\cdot\displaystyle\lim_{h\to 0}\dfrac{v(x+h)-v(x)}{h}+v(x)\cdot\displaystyle\lim_{h\to 0}\dfrac{u(x+h)-u(x)}{h}\\ &=u\dfrac{dv}{dx}+v\dfrac{du}{dx}.\ _\square \end{aligned}$

The product rule can also be applied to the product of numerous variables. For example,

$\frac{ d}{dx} uvw = uv \frac{d}{dx} w + uw \frac{d}{dx} v + vw \frac{d}{dx} u,$

or in Lagrange notation

$(uvw)' = uvw' +uwv' + vwu'.$

What is $\frac{d}{dx} x e^x?$

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Applying the product rule with $u = x$ and $v = e^x$, we get that

$\frac{d}{dx} x e^x = x \frac{d}{dx} e^x + e^x \frac{d}{dx} x = x \times e^x + e^x \times 1 = (x+1) e^x .$

Hence, $\frac{d}{dx} x e^x = (x+1) e^x$. $_\square$

Find the derivative of

$f(x) = x^2\sin x.$

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Since $(x^2)' = 2x$ and $(\sin x)' = \cos x,$

$f'(x) = (2x)(\sin x)+(\cos x)(x^2) = 2x\sin x + x^2\cos x.\ _\square$

What is $\frac{ d}{dx} c f(x)?$

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We may recognize this as a basic property of integerals. Applying the product rule with $u = c$ and $v = f(x)$, we get that

$\frac{d}{dx} c f(x) = c \frac{d}{dx} f(x) + f(x) \frac{d}{dx} c = c f'(x) + f(x) \times 0 = c f'(x).$

Hence, $\frac{d}{dx} c f(x) = c f'(x)$. $_\square$

Some functions may require the combined use of differentiation rules, such as this one here:

If you are having some trouble, you may want to review the chain rule.