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 \(sec^2x\)}
