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 $$sec^2x$$}