Math, asked by roysandip265gmailcom, 1 month ago

Find derivation of : (x^2+3x^3+ 4x^2+2)/x^3

Answers

Answered by Anonymous
3

Given

\to\sf \dfrac{x^2+3x^3+4x^2+2}{x^ 3}  \: differentiate\: with\:respect\:to\:x

We can write as

\to\sf d\dfrac{\bigg(\dfrac{x^2+3x^3+4x^2+2}{x^3}\bigg) }{dx}

Now Simplify the Equation

\to\sf \dfrac{x^2+3x^3+4x^2+2}{x^3} =\dfrac{x^2}{x^3} +\dfrac{3x^3}{x^3} +\dfrac{4x^2}{x^3} +\dfrac{2}{x^3}

\to\sf\dfrac{1}{x} +\dfrac{3}{1} +\dfrac{4}{x} +\dfrac{2}{x^3}

\to\sf x^{-1}+3+4x^{-1}+2x^{-3}

\to\sf \dfrac{d(\sf x^{-1}+3+4x^{-1}+2x^{-3})}{dx}

\to\sf \dfrac{d(x^{-1})}{dx} +\dfrac{d(3)}{dx} +\dfrac{d(4x^{-1})}{dx} +\dfrac{d(2x^{-3})}{dx}

\to\sf\dfrac{d(x^{-1})}{dx} +\dfrac{d(3)}{dx} +4\dfrac{d(x^{-1})}{dx} +2\dfrac{d(x^{-3})}{dx}

\to\sf -1x^{-1-1}+0+4\times-1x^{-1-1}+2\times-3x^{-3-1}

\to\sf -1x^{-2}-4x^{-2} -6x^{-4}

\to\sf\dfrac{-1}{x^2} -\dfrac{4}{x^2} -\dfrac{6}{x^4}

Answer

\to\sf\dfrac{-1}{x^2} -\dfrac{4}{x^2} -\dfrac{6}{x^4}

Answered by arunabalamohapatra
1

Answer:

 \frac{ {x}^{2} + 3 {x}^{3}  + 4 {x}^{2}  + 2 }{ {x}^{3} }  \\  \\ we \: can \: write \: as \\  : \implies \: d  \frac{ \frac{ {x}^{2} + 3 {x}^{3}  + 4 {x}^{2} + 2  }{ {x}^{3} } }{dx}  \\  \\ simplify \:  \: the \:  \: eqation \\  :  \implies \:  \frac{ {x}^{2}  + 3 {x}^{3}  + 4 {x}^{2} + 2 }{ {x}^{3} }  =  \frac{ {x}^{2} }{ {x}^{3} }  +  \frac{3 {x}^{3} }{ {x}^{3} }  +  \frac{4 {x}^{2} }{ {x}^{3} }  +  \frac{2}{ {x}^{3} }  \\  \\ :  \implies \:  \frac{1}{x}  +  \frac{3}{1}  +  \frac{4}{x}  +  \frac{2}{ {x}^{3} }   \\  \\   :  \implies \:  {x}^{ - 1}  + 3 + 4 {x}^{ - 1}  + 2 {x}^{ - 3 }  \\  \\  :  \implies \:   \frac{d( {x}^{ - 1}  + 3  \:  + 4 {x}^{ - 1}  + 2 {x}^{ - 3}) }{dx}  \\  \\  : \implies \frac{d( {x}^{ - 1} )}{dx}  +  \frac{d(3)}{dx}  +  \frac{d(4 {x}^{ - 1} )}{dx}  +  \frac{d(2 {x}^{ - 3}) }{dx}  \\  \\  :  \implies \:  \frac{d( {x}^{ - 1}) }{dx}  +  \frac{d(3)}{dx}  + 4 \frac{d( {x}^{ - 1} )}{dx}  + 2 \frac{d( {x}^{ - 3} )}{dx}  \\  \\  :  \implies - 1x {}^{ - 1 - 1}  + 0 + 4 \times  - 1x {}^{ - 1 - 1}  + 2 \times  - 3x {}^{ - 3 - 1}  \\ \\   : \implies - 1x {}^{ - 2}  - 4x {}^{ - 2}  - 6x {}^{ - 4}  \\  \\  :  \implies \frac{ - 1}{ {x}^{2} }  -  \frac{4}{ {x}^{2} }  -  \frac{6}{ {x}^{4} }

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