# Product Rule

• The product rule is a formula used to find the derivatives of products of two or more functions.

Let $$u\left( x \right)$$ and $$v\left( x \right)$$ be differentiable functions. Then the product of the functions $$u\left( x \right)v\left( x \right)$$ is also differentiable and

${\left( {uv} \right)^\prime } = u’v + uv’.$

We prove the above formula using the definition of the derivative. For this we find the increment of the functions $${uv}$$ assuming that the argument changes by $$\Delta x:$$

${\Delta \left( {uv} \right) }={ u\left( {x + \Delta x} \right)v\left( {x + \Delta x} \right) }-{ u\left( x \right)v\left( x \right).}$

Take into account that

${u\left( {x + \Delta x} \right) = u\left( x \right) + \Delta u,\;\;\;}\kern0pt {v\left( {x + \Delta x} \right) = v\left( x \right) + \Delta v,}$

where $$\Delta u$$ and $$\Delta v$$ are the increments, respectively, of the functions $$u$$ and $$v$$. Omitting for brevity the argument $$x$$ of the functions $$u$$ and $$v$$, we can write the increment $$\Delta \left( {uv} \right)$$ in the following form:

$\require{cancel} {\Delta \left( {uv} \right) = \left( {u + \Delta u} \right)\left( {v + \Delta v} \right) – uv } = {\cancel{uv} + u\Delta v + v\Delta u + \Delta u\Delta v – \cancel{uv} } = {u\Delta v + v\Delta u + \Delta u\Delta v.}$

We proceed to calculate the derivative of the product using the properties of limits

${{\left( {uv} \right)^\prime } = {\lim\limits_{\Delta x \to 0} \frac{{\Delta \left( {uv} \right)}}{{\Delta x}}} } = {\lim\limits_{\Delta x \to 0} \frac{{u\Delta v + v\Delta u + \Delta u\Delta v}}{{\Delta x}} } = {\lim\limits_{\Delta x \to 0} \frac{{u\Delta v}}{{\Delta x}} + \lim\limits_{\Delta x \to 0} \frac{{v\Delta u}}{{\Delta x}} }+{ \lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \Delta v.}$

In the first limit the function $$u$$ does not depend on the increment $$\Delta x$$. Therefore, it can be taken outside the limit sign. The same applies to the function $$v$$ in the second term. We calculate separately the limit $$\lim\limits_{\Delta x \to 0} \Delta v:$$

${\lim\limits_{\Delta x \to 0} \Delta v }={ \lim\limits_{\Delta x \to 0} \left( {\frac{{\Delta v}}{{\Delta x}} \cdot \Delta x} \right) } = {\lim\limits_{\Delta x \to 0} \frac{{\Delta v}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \Delta x }={ v’ \cdot 0 = 0.}$

Thus, the derivative of the product is given by

${{\left( {uv} \right)^\prime } = {\lim\limits_{\Delta x \to 0} \frac{{u\Delta v}}{{\Delta x}} + \lim\limits_{\Delta x \to 0} \frac{{v\Delta u}}{{\Delta x}} }+{ \lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \Delta v }} = {u\lim\limits_{\Delta x \to 0} \frac{{\Delta v}}{{\Delta x}} + v\lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} }+{ \lim\limits_{\Delta x \to 0} \frac{{\Delta u}}{{\Delta x}} \cdot \lim\limits_{\Delta x \to 0} \Delta v } = {uv’ + vu’ + u’ \cdot 0 } = {u’v + uv’.}$

Important: The derivative of the product is NOT equal to the product of the derivatives!

From this formula, it is easy to obtain an expression for the derivative of the function $$kf\left( x \right)$$, where $$k$$ is a constant:

${{\left( {kf\left( x \right)} \right)^\prime } = {k’f\left( x \right) + kf’\left( x \right) }} = {0 \cdot f\left( x \right) + kf’\left( x \right) } = {kf’\left( x \right),}$

so that a constant factor can be taken out of the sign of derivative.

• ## Solved Problems

Click a problem to see the solution.

### Example 1

Let $$y = {\sin ^2}x$$. Differentiate this function without using the chain rule.

### Example 2

Find the derivative of the function $$y = {e^x}\cos x.$$

### Example 3

Find the derivative of the function $$y = {x^2}\sin x.$$

### Example 4

Find the derivative of the function $${y = \sqrt x \left( {1 + x} \right).}$$

### Example 5

Find the derivative of the function $$y = \left( {1 – 2x} \right)\left( {2 – x} \right).$$

### Example 6

Differentiate the function $$y = \left( {3 – 2x} \right)\left( {2 – 3x} \right).$$

### Example 7

Given the function $$z = \left( {{x^2} + 1} \right)\left( {x – 1} \right)$$. Find the value of its derivative at $$x = -1.$$

### Example 8

Determine the derivative of the function $$y = x\ln x – x.$$

### Example 9

Differentiate the function $$y = \left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right).$$

### Example 10

Find the derivative of the fractional function $$y = \large{\frac{{2x + 1}}{{{x^3}}}}\normalsize.$$

### Example 11

Differentiate the function $$y = {x^3}{2^x}.$$

### Example 12

Find the derivative of the function $$y = \left( {{e^x} – 1} \right)\left( {{e^x} + 2} \right)$$ and calculate its value at $$x = 0.$$

### Example 13

Find the derivative of the function $$y = {e^x}\sin x$$ at $$x = 0.$$

### Example 14

Differentiate the function $$y = {e^x}\left( {\sin x – \cos x} \right).$$

### Example 15

Derive a formula for the derivative of the product of three functions.

### Example 16

Find the derivative of the function $$y = \left( {x – 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)$$ at $$x = 0.$$

### Example 1.

Let $$y = {\sin ^2}x$$. Differentiate this function without using the chain rule.

Solution.

We represent the function in the form $$y\left( x \right) = \sin x\sin x$$. By the product rule,

${y’\left( x \right) = {\left( {\sin x\sin x} \right)^\prime } } = {{\left( {\sin x} \right)^\prime }\sin x + \sin x{\left( {\sin x} \right)^\prime }.}$

Since $${\left( {\sin x} \right)^\prime } = \cos x$$, we obtain

${y’\left( x \right) }={ \cos x\sin x + \sin x\cos x } = {2\sin x\cos x = \sin 2x.}$

### Example 2.

Find the derivative of the function $$y = {e^x}\cos x.$$

Solution.

Differentiating this function as a product, we find:

${y’\left( x \right) = {\left( {{e^x}\cos x} \right)^\prime } } = {{\left( {{e^x}} \right)^\prime }\cos x + {e^x}{\left( {\cos x} \right)^\prime } } = {{e^x}\cos x + {e^x}\left( { – \sin x} \right) } = {{e^x}\left( {\cos x – \sin x} \right).}$

### Example 3.

Find the derivative of the function $$y = {x^2}\sin x.$$

Solution.

By the product rule we obtain:

${y’\left( x \right) }={ {\left( {{x^2}\sin x} \right)^\prime } } = {{\left( {{x^2}} \right)^\prime }\sin x + {x^2}{\left( {\sin x} \right)^\prime } } = {2x\sin x + {x^2}\cos x } = {x\left( {2\sin x + x\cos x} \right).}$

### Example 4.

Find the derivative of the function $${y = \sqrt x \left( {1 + x} \right).}$$

Solution.

Let $$u = \sqrt x ,$$ $$v = 1 + x.$$

Then using the product rule $$\left( {uv} \right)^\prime = u^\prime v + uv^\prime,$$ we have

${y^\prime = \left( {\sqrt x \left( {1 + x} \right)} \right)^\prime }={ {\left( {\sqrt x } \right)^\prime}\left( {1 + x} \right) + \sqrt x {\left( {1 + x} \right)^\prime }}={{\frac{1}{{2\sqrt x }}} \cdot \left( {1 + x} \right) + \sqrt x \cdot {1} }={ \frac{{1 + x}}{{2\sqrt x }} + \sqrt x }={ \frac{{1 + x}}{{2\sqrt x }} + \frac{{2\sqrt x \sqrt x }}{{2\sqrt x }} }={ \frac{{1 + x + 2x}}{{2\sqrt x }} }={ \frac{{1 + 3x}}{{2\sqrt x }}.}$

### Example 5.

Find the derivative of the function $$y = \left( {1 – 2x} \right)\left( {2 – x} \right).$$

Solution.

By the product rule,

${y^\prime }={ \left( {1 – 2x} \right)^\prime\left( {2 – x} \right) }+{ \left( {1 – 2x} \right)\left( {2 – x} \right)^\prime }={ – 2 \cdot \left( {2 – x} \right) }+{ \left( {1 – 2x} \right) \cdot \left( { – 1} \right) }={ – \color{blue}{4} + \color{red}{2x} + \color{blue}{1} + \color{red}{2x} }={ \color{red}{4x} – \color{blue}{3}.}$

### Example 6.

Differentiate the function $$y = \left( {3 – 2x} \right)\left( {2 – 3x} \right).$$

Solution.

${y’\left( x \right) = {\left[ {\left( {3 – 2x} \right)\left( {2 – 3x} \right)} \right]^\prime } } = {{\left( {3 – 2x} \right)^\prime }\left( {2 – 3x} \right) }+{ \left( {3 – 2x} \right){\left( {2 – 3x} \right)^\prime } } = {- 2\left( {2 – 3x} \right) + \left( {3 – 2x} \right)\left( { – 3} \right) } = {-\color{red}{4} + \color{blue}{6x} -\color{red}{9} + \color{blue}{6x} = \color{blue}{12x} – \color{red}{13}.}$

Page 1
Problems 1-6
Page 2
Problems 7-16