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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.

Video tutorial – using the product rule

Watch this video to learn how to use the product rule to find the derivative of a function.

Hi, I’m Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on the product rule.

The product rule is used when we want to differentiate the product of two functions. So, if f of x, is equal to u of x times v of x, then the derivative f dash x is u of x times v dash x plus u dash x times vx. This formula can be abbreviated to y dash is equal to u times v dash, plus v times u dash. Note also that f dash x and y dash mean the same thing. So lets do a couple of examples to illustrate the product rule. Here’s a function, x plus 3 to the power of six multiplied by 2x minus 1. Notice here that there are two functions so we can use the product rule. So, let u equal x plus 3 to the power of 6, therefore the derivative is u dash equals 6 brackets x plus 3 to the power of five and let v equals 2x minus 1, and the derivative there is v dash equals 2. Notice I’ve used the chain rule to work out u dash. Substituting in to the formula y dash equals uv dash plus vu dash I get the following four lines of working. Notice when you get down to the third line that I’ve taken a factor of 2 brackets x plus 3 to the power of 5 out, and in the last line I’ve simplified the expression inside the square brackets to give us an answer of 14x brackets x plus 3 all to the power of 5.

Here’s another example. Again notice that there are two functions here e to the x multiplied by sin of 2x. So, this lends itself to the product rule. So I’m letting u equals e to the x, the derivative is the same e to the x and v equals sin of 2 x from which the derivative is 2 cos of 2 x using the chain rule. Substituting as before into the formula y dash equals uv dash plus vu dash I get the following three lines of working. Notice again always look for terms that will factorise and I’ve done that in the last line. So my answer is y dash is equal to ex brackets 2 cos 2x plus sin of 2 x.

Now try some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.

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. \(\dfrac{e^{x}}{x}-\dfrac{e^{x}}{x^{2}}=e^{x}\left(\dfrac{1}{x}-\dfrac{1}{x^{2}}\right)\)

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