Question

Find derivative of $\left( {x\,sinx} \right)$ with respect to $x$.

Answer 1

We use the product formula of derivatives to find an answer. According to this formula, we will take two variables and operate only the first variable with derivative.

Formula used: $\dfrac{d}{{dx}}(u.v) = \dfrac{{du}}{{dx}}(v) + u\dfrac{{dv}}{{dx}}$
$\dfrac{{d(x)}}{{dx}} = 1,\,\dfrac{d}{{dx}}\sin x = \cos (x)$

Complete step by step answer:
(1) Let $y = x{\text{ }}sin{\text{ }}x$
(2) On comparing,$y = u.v$, we have $u = x,v = \sin x$
(3) Using formula $\dfrac{{dy}}{{dx}} = \dfrac{{du}}{{dx}}(v) + u\dfrac{{dv}}{{dx}}$
Where $u = x,\,\,v = \sin x$
Substituting values of u and v in above, we have
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(x).\,(\sin x) + x\dfrac{d}{{dx}}(\sin x)$ $....(1)$
(4) As we know that for differentiation we have formula of algebraic function
i.e. $y = {(x)^n}$
$\Rightarrow \dfrac{{dy}}{{dx}} = n{(x)^{n - 1}}\dfrac{d}{{dx}}(base)$
And for trigonometric function
$y = \sin A$
Taking derivative, we get
$\dfrac{{dy}}{{dx}} = \cos A\dfrac{d}{{dx}}(A)$
Using this formula in equation $(1)$, we have
$\dfrac{{dy}}{{dx}} = (1)\sin x + (x).\cos x$
$\Rightarrow \dfrac{{dy}}{{dx}} = \sin + x\cos x$
Which is the required derivative of x sinx w.r.t. x.

Note: A derivative is a contract between two parties which derives its value or price from an underlying asset. The most common types of derivatives are futures, options, forwards and swaps.