Calculus

Differentiation of Functions

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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.

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

See more problems on Page 2.

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