Question

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

Answer 1

Hint: 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.