Product Rule

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

In Leibniz’s notation we can express it as

OR

In Lagrange’s notation as,

\((u . v)’ = u’ v + u v’\)

This rule can be extended to a derivative of three or more functions.

History

Gottfried Leibniz is credited with the discovery of this rule, he demonstrated it using differentials.

Derivation

Leibniz’s Argument:

 Let \(u(x)\) and \(v(x)\) be two differentiable functions of x.

Then the differential of \(u.v\) is given by

\(d(u . v) = (u + du) . (v + dv) – u.v\)
\( = udv + vdu + du.dv\)

Since the term \(du.dv\) is negligible compared to \(udv + vdu\), as it becomes very small. So we neglect the term

Hence, Leibniz concluded that

\(d(u . v) = v . du + u . dv\)

dividing bot sides we come to

\(\frac{d}{dx}(u . v) = \frac{du}{dx} . v + u . \frac{dv}{dx}\)

Proof

Let us consider two differentiable functions \(f(x)\) & \(g(x)\), & \(h(x) = f(x)g(x)\).

Here, we will be proving that \(h\) is differential with \(x\) & \(h’(x)\) will be \(f’(x)g(x) + f(x)g’(x)\).

Factorising from first principle,

\(h(x) = f(x)g(x)\)

\(h’(x) = lim_{x \to 0} \frac{h(x + \Delta x) – h(x)}{\Delta x}\)

\(= lim_{x \to 0} \frac{f(x + \Delta x)g(x + \Delta x) – f(x)g(x)}{ \Delta x}\)

Adding and subtracting \(f(x)g(x + \Delta x)\) in numerator,

\(= lim_{x \to 0} \frac{ f(x + \Delta x)g(x + \Delta x) –f(x)g(x + \Delta x) + f(x)g(x + \Delta x) – f(x)g(x)}{ \Delta x}\)

\(= lim_{x \to 0} \frac{[ f(x + \Delta x) + f(x)] . g(x + \Delta x) + f(x) . [g(x + \Delta x) + g(x)]}{\Delta x}\)

\(= lim_{x \to 0} \frac{f(x + \Delta x) + f(x)}{\Delta x} . lim_{x \to 0} g(x + \Delta x) + lim_{x \to 0} f(x). lim_{x \to 0} \frac{g(x + \Delta x) + g(x)}{\Delta x}\)

Since, \(lim_{x \to 0} g(x + \Delta x) = g(x)\)

\(h’(x) = f’(x)g(x) + f(x)g’(x)\)

Hence proved.

Application of Product Rule 

This rule is used mainly in calculus and is important when one has to differentiate product of two or more functions. It makes calculation clean and easier to solve.

Examples

Question 1. Differentiate the function: \((x^3 + 5)(x^2 + 1)\)

Solution. Here, \(f(x) = (x^3 + 5)\) & \(g(x) = (x^2 + 1)\)

Using this rule we get,

\(\frac{d (x^3 + 5) (x^2 + 1)}{dx} = \frac{d (x^3 + 5)}{dx}.(x^2 + 1) + (x^3 + 5). \frac{ d (x^2 + 1)}{dx}\)

= \((3x^2)(x^2 +1) + (x^3 + 5)(2x)\) = \(3x^4 + 3x^2 + 2x^4 + 10x\)

=> \(5x^4 + 3x^2 + 10x\).

Question 2. Find the derivative of \(h(x)\), if \(h(x) = f(x)g(x)\) & \(f(x) = sin(x)\) & \(g(x) = cos(x)\).

Solution. Given, \(h(x) = sin(x) . cos(x)\)

Therefore, \(h’(x) = cos(x)cos(x) + sin(x)(- sin(x))\)

\(\Rightarrow cos^2(x) – sin^2(x)\)

\(\Rightarrow cos(2x)\).

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