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The product rule

The product rule helps us differentiate functions that are one function multiplied by another. Use this resource to learn how to apply the product rule.

The product rule for differentiation

The product rule states that if \(y=f(x)=u(x)\cdot v(x)\), then:

\[\frac{dy}{dx} = u(x)\cdot\frac{dv}{dx}+v(x)\cdot\frac{du}{dx}\]

You may see this written as \(y'=uv'+u'v\).

Example 1 – differentiating using the product rule

Find the derivative of \(f(x)=(x+3)^{6}(2x-1)\).

Let \(u=(x+3)^{6}\) and \(v=2x-1\). For \(u\), we need to use the chain rule, so let's do that first. We can let \(w=x+3\) and \(u=w^{6}\).
\[\begin{align*} u' & = \frac{du}{dw}\times\frac{dw}{dx}\\
& = 6w^{5}\times1\quad\textrm{substitute in }w=x+3\\
& = 6(x+3)^{5}
\end{align*}\]

Now, we can use the product rule.
\[\begin{align*} f'(x) & = u\cdot v' + u'\cdot v\\
& = ((x+3)^{6}\cdot 2) + (6(x+3)^{5}\cdot(2x-1)\\
& = (x+3)^{5}\left(2(x+3)+6(2x-1)\right)\\
& = 14x(x+3)^{5}
\end{align*}\]

Differentiate \(e^{x}\sin(2x)\).

Let \(u=e^{x}\) and \(v=\sin(2x)\). We can let \(y=e^{x}\sin(2x)\). Using the product rule:
\[\begin{align*} y' & = uv'+u'v\\
& = (e^{x}\cdot2\cos(2x)) + (e^{x}\cdot\sin(2x))\\
& = 2e^{x}\cos(2x)+e^{x}\sin(2x)\\
& = e^{2}(2\cos(2x)+\sin(2x))
\end{align*}\]

Exercise – differentiating using the product rule

  1. Use the product rule to differentiate the following.
    1. \(y=(x-2)(6x+7)\)
    2. \(f(x)=(2x^{2}+4)(x^{5}+4x^{2}-2)\)
    3. \(y=(\sqrt{x}-1)(x^{2}+1)\)
    4. \(y=(x^{3}-4x+\sqrt{x})(3x^{4}+2)\)
  2. Find the derivative of the following.
    1. \(y=e^{x}\tan(x)\)
    2. \(y=x{^2}\log_{e}(x)\)
    3. \(y=\sin(x)\cos(x)\)
    4. \(y=\dfrac{e^{x}}{x}\)

    1. \(12x-5\)
    2. \((2x^{2}+4)(5x^{4}+8x)+4x(x^{5}+4x^{2}-2)\)
    3. \(\dfrac{5}{2}x^{\frac{3}{2}}-2x+\dfrac{1}{2\sqrt{x}}\)
    4. \(12x^{3}(x^{3}-4x+\sqrt{x})+(3x^{4}+2)\left(3x^{2}-4+\dfrac{1}{2\sqrt{x}}\right)\)
    1. \(e^{x}\tan(x)+e^{x}\sec^{2}(x)\)
    2. \(x+2x\log_{e}(x)\)
    3. \(\cos^{2}(x)-\sin^{2}(x)\)
    4. \(\dfrac{e^{x}}{x}-\dfrac{e^{x}}{x^{2}}=e^{x}\left(\dfrac{1}{x}-\dfrac{1}{x^{2}}\right)\)

The product rule in context

In robotics, a start-up company, Mecha Methods, is designing a mechanical arm that can extend and retract smoothly with precision. The gear controlling the arm’s motion follows the function \(f(x)=3x^{2}(5x+4)^{3}\), where \(x\) represents the angle at which the arm operates.

Differentiate the function with respect to \(x\) to determine how the mechanical force varies with the angle of operation.

Let \(y=uv\) where \(u=3x^{2}\) and \(v=(5x+4)^{3}\).
\[\frac{du}{dx}=6x\]

Using the chain rule:
\[\frac{dv}{dx}=15(5x+4)^{2}\]

Using the product rule:
\[\begin{align*} \frac{dy}{dx} & = u(x)\cdot\frac{dv}{dx}+v(x)\cdot\frac{du}{dx}\\
& = 45x^{2}(5x+4)^{2}+6x(5x+4)^{3}\\
& = [2(5x+4)+15x]3x(5x+4)^{2}\\
& = 3x(5x+4)^{2}(25x+8)
\end{align*}\]

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