Product Rule

The rule

To differentiate a product of two functions, differentiate each one in turn while holding the other fixed, then add the results:

\dfrac{d}{dx}\left[u(x)v(x)\right] = u'(x)v(x) + u(x)v'(x)

Find the derivative of $f(x) = (x^2+1)(x^3-3x)$ .

Solution: With $u=x^2+1$ (so $u'=2x$ ) and $v=x^3-3x$ (so $v'=3x^2-3$ ):

f'(x) = 2x(x^3-3x) + (x^2+1)(3x^2-3)

= (2x^4-6x^2) + (3x^4-3) = 5x^4 - 6x^2 - 3

The rule extends to any number of factors: differentiate one factor at a time, keep the others unchanged, and add up the results:

\dfrac{d}{dx}\left[uvw\right] = u'vw + uv'w + uvw'

Differentiate $y = x^2 e^x \sin x$ .

Solution: With $u = x^2$ , $v = e^x$ and $w = \sin x$ , the derivatives are $u' = 2x$ , $v' = e^x$ and $w' = \cos x$ :

y' = 2x\,e^x\sin x + x^2e^x\sin x + x^2e^x\cos x = xe^x\left(2\sin x + x\sin x + x\cos x\right)

When a variable appears in both the base and the exponent, taking a logarithm first turns the problem into one the product rule can handle.

If $y = x^{\sin x}$ for $x > 0$ , find $\dfrac{dy}{dx}$ .

Solution: Take the natural log of both sides: $\ln y = \sin x \ln x$ . Differentiate both sides: the right side is now a product:

\dfrac{y'}{y} = \cos x \ln x + \sin x \cdot \dfrac{1}{x}

Multiplying back through by $y = x^{\sin x}$ :

y' = x^{\sin x}\left[\dfrac{\sin x}{x} + \ln x \cos x\right]

Multiplying the derivatives. $(uv)' \ne u'v'$ . Try $u = v = x$ : the product is $x^2$ , whose derivative is $2x$ , but $u'v' = 1$ .
Dropping one of the terms. There is one term per factor: two for $uv$ , three for $uvw$ . Each term has exactly one factor differentiated.
Reaching for it when one factor is a constant. $5x^3$ is a constant times a power, so its derivative is just $15x^2$ . The product rule gives the same answer, only more slowly.
Worrying about the order. Addition doesn't care about order, so $u'v + uv'$ and $uv' + u'v$ are the same. That is not true of the quotient rule, where order matters.