Question

If y = x^2sin2x , then the first derivative is​

Answer 1

$\bf\large \underbrace\red{Answer:}$

Given :-

$\bf \:y = {x}^{2} sin2x$

To find :-

$\bf \:\dfrac{dy}{dx}$

Formula used :-

$\bf \:\dfrac{d}{dx} (u.v) = u\dfrac{d}{dx} v + v\dfrac{d}{dx} u$

$\bf \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}$

$\bf \:\dfrac{d}{dx} sinx = cosx$

$\bf\underbrace\orange{Solution:}$

$\bf \:y = {x}^{2} sin2x$

Differentiate w. r. t. x, we get

$\bf \:\dfrac{dy}{dx} = \dfrac{d}{dx} ( {x}^{2} sin2x)$

$\bf \:\dfrac{dy}{dx} = {x}^{2} \dfrac{d}{dx} sin2x + sin2x\dfrac{d}{dx} {x}^{2}$

$\bf \:\dfrac{dy}{dx} = {x}^{2} cos2x\dfrac{d}{dx} 2x + sin2x \times 2x$

$\bf \:\dfrac{dy}{dx} = 2 {x}^{2} cos2x + 2xsin2x$

$\bf\implies \:\dfrac{dy}{dx} = 2x(xcos2x + sin2x)$

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