Question

x sin x derivates function from the first principle​

Answer 1

Question :-

$\bf{find \: the \: derivative \: of \: x \: \sin( x)} \\ \bf{ by \: \: first \: principal \: }$

Answer :-

$\rightarrow \: \red{ \boxed{\bf{ \frac{ dy}{dx} = x \: \cos(x) + \sin(x) }} \: }$

Formula used :-

Here we used first principal of differentiation,

$\implies \: \small \boxed{ \bf{ \frac{d}{dx} m \: n\: = m \: \frac{d}{dx} n \: + n \: \frac{d}{dx} m}} \\ \\ \implies \: \star \: \bf{ \frac{d}{dx} \sin(x) = \cos(x) }\\ \\ \implies \: \bf{ \star \: \frac{d}{dx} \: x \: = 1 }$

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Solution :-

Let,

$\implies \: \red{\bf{y \: = x \sin(x) }}$

Now differentiating on both sides with respect to "x",

$\implies \: \bf{ \frac{dy}{dx} = \frac{d}{dx} \big(x \: \sin(x) \big) }$

Using the given formula of first principal of differentiation .

$\rightarrow \: \: \small \bf{ \frac{dy}{dx} = x \: \frac{d}{dx} \sin(x) + \sin(x) \frac{d}{dx} \: x } \\ \\ \rightarrow \: \bf{ \frac{dy}{dx} = x \: \cos(x) + \sin(x) \times 1} \\ \\ \rightarrow \: \boxed{\bf{ \frac{ dy}{dx} = x \: \cos(x) + \sin(x) }}$

This is the required solution.

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