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

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]