# The Product Rule for Differentiation

The product rule is the method used to differentiate the product of two functions.
For instance, if we were given the function defined as: $f(x)=x^2sin(x)$ this is the product of two functions, which we typically refer to as $$u(x)$$ and $$v(x)$$.

So, in the case of $$f(x)=x^2sin(x)$$, we would define $$u(x)=x^2$$ and $$v(x)=sin(x)$$, to write: $f(x)=u(x)\times v(x)$

## Product Rule

Given a function that can be written as the product of two functions: $f(x)=u(x).v(x)$ we can differentiate this function using the product rule: $\text{if} \quad f(x)=u(x). v(x)$ $\text{then} \quad f'(x)=u'(x).v(x)+u(x).v'(x)$ This formula is further explained and illustrated, with some worked examples, in the following tutorial.

## Tutorial

In the following tutorial we review the product rule and learn how to use it with some examples.

## Exercise 1

1. Given $$f(x) = 5x.cos(x)$$, find $$f'(x)$$.

2. Given $$y=x^2.ln(x)$$, find an expression for $$\frac{dy}{dx}$$.

3. Given $$y= \begin{pmatrix} 2x^3 - 2x \end{pmatrix} e^x$$, find an expression for $$\frac{dy}{dx}$$.

4. Find $$f'(x)$$ where $$f(x) = 5.ln(x)tan(x)$$.

5. Given $$y= 2 sin(x) \sqrt{x}$$, find an expression for $$\frac{dy}{dx}$$.

6. Given $$f(x) = \sqrt[3]{x}\begin{pmatrix} x^3-2x \end{pmatrix}$$, find $$f'(x)$$.

1. $$f'(x) = 5cos(x)-5xsin(x)$$

2. $$\frac{dy}{dx} = 2x.ln(x)+x$$

3. $$f'(x) = e^x \begin{pmatrix} 2x^3 + 6x^2 - 2x - 2 \end{pmatrix}$$

4. $$f'(x) = \frac{5tan(x)}{x} + 5ln(x).sec^2(x)$$

5. $$\frac{dy}{dx} = 2\sqrt{x}.cos(x) + \frac{sin(x)}{\sqrt{x}}$$

6. $$f'(x) = \frac{10 \sqrt[3]{x}}{3} - \frac{8 \sqrt[3]{x}}{3}$$

## Solution With Working

1. The function $$f(x) = 5x.cos(x)$$ can be written $$f(x) = u(x).v(x)$$.

Where $u(x) = 5x \quad \text{and}\quad v(x) = cos(x)$ and $u'(x) = 5 \quad \text{and} \quad v'(x) = -sin(x)$ Using the product rule we can now find the derivative: \begin{aligned} f'(x) &= u'(x).v(x)+ u(x).v'(x) \\ & = 5.cos(x) + 5x.\begin{pmatrix}-sin(x) \end{pmatrix} \\ f'(x) & = 5cos(x) -5xsin(x) \end{aligned}

2. The function $$y = x^2.ln(x)$$ can be written $$y = u(x).v(x)$$.
Where $u(x) = x^2 \quad \text{and} \quad v(x) = ln(x)$ and $u'(x) = 2x \quad \text{and} \quad v'(x) = \frac{1}{x}$ Using the product rule we can now find the derivative: \begin{aligned} \frac{dy}{dx} & = u'(x).v(x) + u(x).v'(x) \\ & = 2x.ln(x) + x^2.\frac{1}{x} \\ & = 2x.ln(x) + \frac{x^2}{x} \\ \frac{dy}{dx} & = 2x.ln(x) + x \end{aligned}

3. The function $$y = \begin{pmatrix} 2x^3 - 2x \end{pmatrix} e^x$$ can be written $$y = u(x).v(x)$$.
Where $u(x) = 2x^3 - 2x \quad \text{and} \quad v(x) = e^x$ and $u'(x) = 6x^2 - 2 \quad \text{and} \quad v'(x) = e^x$ Using the product rule we can now find the derivative: \begin{aligned} f'(x) & = u'(x).v(x) + u(x).v'(x) \\ & = \begin{pmatrix} 6x^2 - 2 \end{pmatrix} e^x + \begin{pmatrix} 2x^3 - 2x \end{pmatrix} e^x \\ & = e^x \begin{pmatrix} 6x^2 - 2 + 2x^3 - 2x \end{pmatrix} \\ f'(x) &= e^x \begin{pmatrix} 2x^3+6x^2 -2x - 2 \end{pmatrix} \end{aligned}

4. We can write $$f(x) = 5ln(x)tan(x)$$ as $$f(x) = u(x).v(x)$$.
Where $u(x) = 5ln(x) \quad \text{and} \quad v(x) = tan(x)$ and $u'(x) = \frac{5}{x} \quad \text{and} \quad v'(x) = sec^2(x)$ Using the product rule we can now find the derivative: \begin{aligned} f'(x) & = u'(x).v(x) + u(x).v'(x) \\ & = \frac{5}{x}.tan(x) + 5ln(x).sec^2(x)\\ f'(x) & =\frac{5tan(x)}{x} + 5ln(x).sec^2(x) \end{aligned}

5. The function $$y=2sin(x)\sqrt{x}$$ can be written $$y = u(x).v(x)$$, where: $u(x) = 2sin(x) \quad \text{and} \quad v(x) = \sqrt{x}$ and $u'(x) = 2cos(x) \quad \text{and} \quad v'(x) = \frac{1}{2\sqrt{x}}$ Using the product rule we can find the derivative: \begin{aligned} \frac{dy}{dx} & = u'(x).v(x) + u(x).v'(x) \\ & = 2cos(x).\sqrt{x} + 2sin(x).\frac{1}{2\sqrt{x}} \\ & = 2cos(x)\sqrt{x} + \frac{2sin(x)}{2\sqrt{x}} \\ \frac{dy}{dx} & = 2\sqrt{x}cos(x) + \frac{sin(x)}{\sqrt{x}} \end{aligned}

6. The function $$f(x) = \sqrt[3]{x} \begin{pmatrix} x^3 - 2x \end{pmatrix}$$ can be written $$f(x) = u(x).v(x)$$, where: $u(x) = \sqrt[3]{x} \quad \text{and} \quad v(x) = x^3 - 2x$ and $u'(x) = \frac{1}{3\sqrt[3]{x^2}} \quad \text{and} \quad v'(x) = 3x^2 - 2$ Using the product rule we can find the derivative: \begin{aligned} f'(x) & = u'(x).v(x) + u(x).v'(x) \\ & = \frac{1}{3\sqrt[3]{x^2}} \begin{pmatrix} x^3 - 2x \end{pmatrix} + \sqrt[3]{x} \begin{pmatrix} 3x^2 - 2 \end{pmatrix} \\ & = \frac{x^3 - 2x}{3\sqrt[3]{x^2}} + \sqrt[3]{x} .3x^2 - \sqrt[3]{x}.2 \\ & = \frac{x^3}{3\sqrt[3]{x^2}} - \frac{2x}{3\sqrt[3]{x^2}} + 3x^{\frac{1}{3}} .x^2 - 2\sqrt[3]{x} \\ & = \frac{x^3}{3x^{\frac{2}{3}}} - \frac{2x}{3x^{\frac{2}{3}}} + 3x^{\frac{1}{3}+2} - 2\sqrt[3]{x} \\ & = \frac{x^3.x^{-\frac{2}{3}}}{3} - \frac{2x.x^{-\frac{2}{3}}}{3} + 3x^{\frac{7}{3}} - 2\sqrt[3]{x} \\ & = \frac{x^{3-\frac{2}{3}}}{3} - \frac{2x^{1-\frac{2}{3}}}{3} + 3\sqrt[3]{x^7} - 2\sqrt[3]{x} \\ & = \frac{x^{\frac{7}{3}}}{3} - \frac{2x^{\frac{1}{3}}}{3} + 3\sqrt[3]{x^7} - 2\sqrt[3]{x} \\ & = \frac{\sqrt[3]{x^7}}{3} - \frac{2\sqrt[3]{x}}{3} + 3\sqrt[3]{x^7} - 2\sqrt[3]{x} \\ f'(x) & = \frac{10\sqrt[3]{x^7}}{3} - \frac{8\sqrt[3]{x}}{3} \end{aligned}

## Exercise 2

Consider the curve defined by $$y=2x.ln(x)$$.

1. Calculate the height of this curve when $$x = 1$$.

2. Find an expression for $$\frac{dy}{dx}$$.

3. Calculate the gradient of the curve when $$x = 1$$.

4. Find the equation of the tangent to the curve, at the point with $$x$$-coordinate $$x = 1$$.

5. Find the equation of the tangent to the curve, at the point with $$x$$-coordinate $$x = 1$$.

## Solution Without Working

1. $$y=0$$

2. $$\frac{dy}{dx} = 2.ln(x)+2$$

3. $$\frac{dy}{dx} = 2$$

4. $$y = 2x - 2$$

5. $$y = - \frac{x}{2} + \frac{1}{2}$$

## Solution With Working

1. The height of the curve, at any value of $$x$$, is calculated with its equation $$y=2x.ln(x)$$.
When $$x = 1$$ that's: \begin{aligned} y & = 2 \times 1 \times ln(1) \\ & = 2 \times 1 \times 0 \\ y & = 0 \end{aligned}

2. To find the derivative, $$\frac{dy}{dx}$$, we use the product rule.
Let $u(x) = 2x \quad \text{and} \quad v(x) = ln(x)$ Then $u'(x) = 2 \quad \text{and} \quad v'(x) = \frac{1}{x}$ Now, using the product rule, we find the derivative: \begin{aligned} \frac{dy}{dx}&= u'(x).v(x) + u(x).v'(x) \\ & = 2.ln(x)+ 2x.\frac{1}{x} \\ & = 2ln(x)+\frac{2x}{x} \\ \frac{dy}{dx} &= 2ln(x) + 2 \end{aligned}

3. The gradient of the curve, when $$x=1$$, equals to the value of $$\frac{dy}{dx} = 2ln(x) + 2$$ when $$x = 1$$.
That's: \begin{aligned} \frac{dy}{dx} &= 2 .ln(1) + 2 \\ & = 2 \times 0 + 2 \\ \frac{dy}{dx} &= 2\end{aligned}

4. To find the tangent to the curve when $$x = 1$$, we follow our two-step method:

• Step 1: find the gradient of the curve when $$x = 1$$.
This was done in question 3). The gradient is: $m = 2$
• Step 2: rearrange the formula $$y - b = m \begin{pmatrix} x - a \end{pmatrix}$$, where $$a$$ and $$b$$ are the $$x$$ and $$y$$ coordinates of the point along the curve.
We know that $$a = 1$$ and in question 1) we found $$b = 0$$. So the formula becomes: $y - 0 = 2 \begin{pmatrix} x - 1 \end{pmatrix}$ This leads to the tangent's equation: $y = 2x - 2$

5. To find the normal to the curve at the same point, we follow our two-step method for normals:
• Step 1: find the normal's gradient, using the fact that is is calculated with the formula: $m = - \frac{1}{\text{gradient of the curve}}$ at the point $$\begin{pmatrix}1,0\end{pmatrix}$$.
We know, from question 3), that the gradient of the curve at this point is $$2$$ so the gradient of the normal is: $m = - \frac{1}{2}$

• Step 2: find the equation of the normal by rearranging the formula $$y-b = m\begin{pmatrix} x - a \end{pmatrix}$$, where $$\begin{pmatrix} a,b \end{pmatrix} = \begin{pmatrix} 1,0 \end{pmatrix}$$ and $$m = - \frac{1}{2}$$.
That's: $y - 0 = - \frac{1}{2} \begin{pmatrix} x - 1 \end{pmatrix}$ This leads us to the equation for the normal: $y = -\frac{x}{2} + \frac{1}{2}$