# 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

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

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 or tap 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 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,

$y^\prime = \left( {1 - 2x} \right)^\prime\left( {2 - x} \right) + \left( {1 - 2x} \right)\left( {2 - x} \right)^\prime = - 2 \cdot \left( {2 - x} \right) + \left( {1 - 2x} \right) \cdot \left( { - 1} \right) = - \color{blue}{4} + \color{red}{2x} - \color{blue}{1} + \color{red}{2x} = \color{red}{4x} - \color{blue}{5}.$

### Example 6.

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

Solution.

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