Product Rule in Differentiation with Examples

Here you will learn what is product rule in differentiation with examples.

Let’s begin –

Product Rule in Differentiation

If f(x) and g(x) are differentiable functions, then f(x)g(x) is also differentiable function such that

\(d\over dx\) {f(x) g(x)} = \(d\over dx\) (f(x)) g(x) + f(x). \(d\over dx\) (g(x))

If f(x), g(x) and h(x) are differentiable functions, then

\(d\over dx\) (f(x) g(x) h(x)) = \(d\over dx\) (f(x)) g(x) h(x) + f(x). \(d\over dx\) (g(x)) h(x) + f(x) g(x) \(d\over dx\) (h(x))  

Example 1 : find the differentiation of sinx cosx.

Solution : Let sinx = f(x) and g(x) = cosx

Then, by using product rule in differentiation,

\(d\over dx\) {f(x) g(x)} = \(d\over dx\) (f(x)) g(x) + f(x). \(d\over dx\) (g(x))

\(d\over dx\) [sinx.cosx] = \(d\over dx\) (sinx) cosx + sinx. \(d\over dx\) (cosx)

= cosx cosx + sinx (-sinx)

= \(cos^2x\) – \(sin^2x\)

= cos2x

Example 2 : find the differentiation of x sinx.

Solution : Let x = f(x) and g(x) = sinx

Then, by using product rule in differentiation,

\(d\over dx\) {f(x) g(x)} = \(d\over dx\) (f(x)) g(x) + f(x). \(d\over dx\) (g(x))

\(d\over dx\) [x.sinx] = \(d\over dx\) (x) sinx + x. \(d\over dx\) (sinx)

= 1.sinx + x.(cosx)

= sinx + x cosx

Example 3 : find the differentiation of \(e^x log\sqrt{x} tanx\).

Solution : Let \(e^x\) = f(x) , g(x) = \(log\sqrt{x}\) and h(x) = tanx

Then, by using product rule,

\(d\over dx\) {f(x) g(x) h(x)} = \(d\over dx\) (f(x)) g(x) h(x) + f(x). \(d\over dx\) (g(x)) h(x) + f(x) g(x) \(d\over dx\) (h(x))

\(d\over dx\) [ \(e^x log\sqrt{x} tanx\)] = \(d\over dx\) [ \(e^x \times {1\over 2} logx \times tanx\)]

= \(1\over 2\) \(d\over dx\) [ \(e^x logx  tanx\)]

= \(1\over 2\) [{\(d\over dx\) (\(e^x\))} logx tanx  + \(e^x\). {\(d\over dx\) (logx)} tanx + \(e^x\) logx {\(d\over dx\) (tanx)}]

= \(1\over 2\) { \(e^x\) logx tanx  + \(e^x\). {\(1\over x\)} tanx + \(e^x\) logx \(sec^2x\)}

= \(1\over 2\) \(e^x\) { logx tanx  + \(tanx\over x\) + logx \(sec^2x\)}

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