How to Find Derivatives using the Product Rule

The product rule is the rule you use when you have a product of two or more functions.

Let $u (x)$ and $v (x)$ be two functions of $x$ . The product rule tells you how to differentiate the product of these two functions. The product is the new function $u (x) \cdot v (x)$ . To simplify it, you just write $u$ and $v$ for the functions $u (x)$ and $v (x)$ , but it’s important to remember that these are functions of $x$ .

Formula

The Product Rule

{(u v)}^{'} = u^{'} v + u v^{'},

where $u = u (x)$ and $v = v (x)$ .

Example 1

Differentiate the expression $x^{2} \sqrt{2 x}$

Looking at the expression, you decide that $u = x^{2}$ and $v = \sqrt{2 x}$ . That gives you $u^{'} = 2 x$ and $v^{'} = \frac{1}{\sqrt{2 x}}$ , and the differentiation is

\begin{array}{l} {(x^{2} \sqrt{2 x})}^{'} & = {(x^{2})}^{'} \cdot \sqrt{2 x} + x^{2} \cdot {(\sqrt{2 x})}^{'} \\ = 2 x \cdot \sqrt{2 x} + x^{2} \cdot \frac{1}{\sqrt{2 x}} \\ = 2 x \sqrt{2 x} + \frac{x^{2}}{\sqrt{2 x}} \\ = \frac{2 x \sqrt{2 x} \cdot \sqrt{2 x}}{\sqrt{2 x}} + \frac{x^{2}}{\sqrt{2 x}} \\ = \frac{4 x^{2}}{\sqrt{2 x}} + \frac{x^{2}}{\sqrt{2 x}} \\ = \frac{5 x^{2}}{\sqrt{2 x}} \end{array}

Example 2

Differentiate the expression $e^{x} (2 x^{3} + 3 x)$

Here, you decide to say that $u = e^{x}$ and $v = 2 x^{3} + 3 x$ . That gives you $u^{'} = e^{x}$ and $v^{'} = 6 x^{2} + 3$ , and the differentiation becomes

\begin{array}{l} {[e^{x} (2 x^{3} + 3 x)]}^{'} & = {(e^{x})}^{'} \cdot (2 x^{3} + 3 x) \\ + e^{x} \cdot {(2 x^{3} + 3 x)}^{'} \\ = e^{x} \cdot (2 x^{3} + 3 x) \\ + e^{x} \cdot (6 x^{2} + 3) \\ = 2 x^{3} e^{x} + 3 x e^{x} + 6 x^{2} e^{x} + 3 e^{x} \\ = e^{x} (2 x^{3} + 6 x^{2} + 3 x + 3) \end{array}

\begin{array}{l} {[e^{x} (2 x^{3} + 3 x)]}^{'} & = {(e^{x})}^{'} \cdot (2 x^{3} + 3 x) + e^{x} \cdot {(2 x^{3} + 3 x)}^{'} \\ = e^{x} \cdot (2 x^{3} + 3 x) + e^{x} \cdot (6 x^{2} + 3) \\ = 2 x^{3} e^{x} + 3 x e^{x} + 6 x^{2} e^{x} + 3 e^{x} \\ = e^{x} (2 x^{3} + 6 x^{2} + 3 x + 3) \end{array}

Example 3

Differentiate the expression $(x^{2} + x - 1) \ln x$

Here, you decide that $u = x^{2} + x - 1$ and $v = \ln x$ . That gives you $u^{'} = 2 x + 1$ and $v^{'} = \frac{1}{x}$ , and the differentiation becomes

\begin{array}{l} {[(x^{2} + x - 1) \ln x]}^{'} & = {(x^{2} + x - 1)}^{'} \cdot \ln x \\ + (x^{2} + x - 1) \cdot {(\ln x)}^{'} \\ = (2 x + 1) \cdot \ln x \\ + (x^{2} + x - 1) \cdot \frac{1}{x} \\ = 2 x \ln x + \ln x + x + 1 - \frac{1}{x} \end{array}

\begin{array}{l} {[(x^{2} + x - 1) \ln x]}^{'} & = {(x^{2} + x - 1)}^{'} \cdot \ln x + (x^{2} + x - 1) \cdot {(\ln x)}^{'} \\ = (2 x + 1) \cdot \ln x + (x^{2} + x - 1) \cdot \frac{1}{x} \\ = 2 x \ln x + \ln x + x + 1 - \frac{1}{x} \end{array}

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