Physics, asked by vickymahato334, 9 months ago

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

Answers

Answered by nirman95
10

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.

Answered by Anonymous
50

 \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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nirman95: Awesome!!
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