Product Rule Of Differentiation – Engineering Mathematics

Suppose is a product of two functions and . This means

Now my task is to differentiate , that is, to get the value of $\cfrac{dy}{dx}.$ Since is a product of two functions, I’ll use the product rule of differentiation to get the value of $\cfrac{dy}{dx}.$ Thus $\cfrac{dy}{dx}$ will be

\[\frac{dy}{dx} = u(x)\frac{dv}{dx}+ v(x)\frac{du}{dx}.\]

Next, I will give some example .

EXAMPLE

According to Stroud and Booth (2013)* “Differentiate $y = x^2 \cos^ 2 x$ .”

SOLUTION

STEP 1

Here the given function is: $y = x^2 \cos^2 x$ .

Now is a product of two functions and $\cos^2 x$ .

In order to differentiate with respect to , I’ll use the product rule of differentiation.

Thus it will be

\[\frac{dy}{dx} = x^2\frac{d}{dx} (\cos^2 x) + \cos^2 x \frac{d}{dx}(x^2).\]

So this means

\[\frac{dy}{dx} = x^2 (2 \cos x)\frac{d}{dx}(\cos x) + \cos^2 x (2x).\]

Now this gives

\[\frac{dy}{dx} = x^2 (2 \cos x)(-\sin x) + 2x \cos^2 x.\]

At the end I’ll simplify it to get the value of $\cfrac{dy}{dx}$ as

\begin{eqnarray*} \frac{dy}{dx} &=& -2 x^2 \cos x \sin x + 2x \cos^2 x\\ &=& 2x \cos x (\cos x - x \sin x). \end{eqnarray*}

Hence I can conclude that this is the answer to the given example.