Question

If y=xsinx^2 then the value of dy/dx is

Answer 1

Answer:

$y = x \sin( {x}^{2} )$

$= > \dfrac{dy}{dx} = \dfrac{dx}{dx} \{\sin( {x}^{2} ) \} + x \dfrac{d \{ \sin( {x}^{2} ) \} }{dx}$

$= > \dfrac{dy}{dx} = \sin( {x}^{2} ) + x \dfrac{d \{ \sin( {x}^{2} ) \}}{d( {x}^{2}) } \times \dfrac{d( {x}^{2}) }{dx}$

$= > \dfrac{dy}{dx} = \sin( {x}^{2} ) + 2 {x}^{2} \cos( {x}^{2} )$

So final answer is :

$\boxed{ \red{ \bold{ \sf{ \dfrac{dy}{dx} = \sin( {x}^{2} ) + 2 {x}^{2} \cos( {x}^{2} ) }}}}$

Things to remember while doing this type of Differentiation are as follows :

• You have to apply product rule while differentiation.
• You also have to apply chain rule while differentiation of complex functions.
• While using chain rule, try to break the components into simple forms and write them.

Answer 2

$\mathtt{ \huge{ \fbox{Solution :)}}}$

Let , f(x) = xSin(x)²

$\star \: \mathtt{ \{Product \: rule \}} \\ \large \mathtt{ \fbox{ \frac{d}{dy}F(x).G(x) =G(x) \frac{d}{dy}F(x) + F(x)\frac{d}{dy} G(x) }} \\ \\ </p><p>\star \: \mathtt{ \{ Chain \: rule </p><p>\} } \\ \large \mathtt{ \fbox{ \frac{d}{dy}F(g(x)) = \frac{d}{dy} F(x) \times \frac{d}{dy}g(x) }} \\ \\ \star \: \mathtt{ \{Power\: rule \}} \\ \large \mathtt{ \fbox{ \frac{d}{y} {(x)}^{n} = n {(x)}^{n - 1} }}$

Using product , chain rule and power rule , we get

$\sf \hookrightarrow \frac{df(x)}{dx} = Sin{(x)}^{2} \frac{d(x)}{dx} + x \frac{d(Sin {(x)}^{2} )}{dx} \\ \\\sf \hookrightarrow \frac{df(x)}{dx} = Sin {(x)}^{2} + x \times \frac{d {(x)}^{2} }{dy} \times \frac{Sin {(x)}^{2} }{dx} \\ \\\sf \hookrightarrow \frac{df(x)}{dx} = Sin {(x)}^{2} + x \times 2{x}^{(2 - 1)} \times Cos{(x)}^{2} \\ \\\sf \hookrightarrow \frac{df(x)}{dx} = Sin {(x)}^{2} + 2 {(x)}^{2} \times Cos{(x)}^{2}$

Hence ,the derivative of xSin(x)² is Sin(x)² + 2(x)² × Cos(x)²

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