Question

Find $\sf{\frac{dy}{dx} }$ when y = $\sf{ \frac{x^{2} - x + 1 }{x^{2} + x + 1}}$

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Answer 1

Answer:

$\bold{ \huge{ \red{ \dfrac{2( {x}^{2} - 1)}{ {( {x}^{2} + x + 1 )}^{2} } }}}$

Step-by-step explanation:

$y = \dfrac{ {x}^{2} - x + 1 }{ {x}^{2} + x + 1 } \\ differentiating \: withrespect \: to \: x \\ \dfrac{dy}{dx} = \dfrac{d}{dx} ( \dfrac{ {x}^{2} - x + 1 }{ {x}^{2} + x + 1 } ) \\ = \dfrac{( {x}^{2} + x + 1) \: \frac{d}{dx} ( {x}^{2} - x + 1) - ( {x}^{2} - x + 1) \: \frac{d}{dx}( {x}^{2} + x + 1) }{ {( {x}^{2} + x + 1) }^{2} } \\ = \dfrac{( {x}^{2} +x + 1) \: (2x - 1) - ( {x}^{2} - x + 1) \: (2x + 1) }{ {( {x}^{2} + x + 1) }^{2} } \\ = \dfrac{2 {x}^{3} + 2 {x}^{2} + 2x - {x}^{2} - x - 1 - 2 {x}^{3} + 2 {x}^{2} - 2x - {x}^{2} + x - 1 }{ ({ {x}^{2} + x + 1) }^{2} } \\ = \dfrac{4 {x}^{2} - 2 {x}^{2} - 2}{{( {x}^{2} + x + 1)} ^{2} } \\ = \dfrac{2 {x}^{2} - 2}{ {( {x}^{2} + x + 1 )}^{2} } \\ = \dfrac{2( {x}^{2} - 1) }{ {( {x}^{2} + x + 1)}^{2} }$

$\large{ \bold{ \pink{ \underline{ \underline{formula \: used}}}}} \\ 1. \blue{ \bold{\dfrac{d}{dx} ( \dfrac{u}{v}) = \dfrac{v \: \: \dfrac{du}{dx} - u \: \: \dfrac{dv}{dx} }{ {v}^{2} } }} \\ \\ \\ 2. \bold{ \green{\dfrac{d}{dx} ( {x}^{n} ) = n \: {x}^{n - 1} }}$