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 $sec^2x$}