Question

of

y = (x^4+x^2+1)/(x^2-x+1)

then

dy/dx = ???

Answer 1

y = ( x^4 + x^2 + 1 ) / (x² - x + 1 )

y ( x² - x + 1 ) = x^4 + x² + 1

Now, Differentiating with respect to x

We know that,

d(uv)/dx = v du/dx + u dv/dx

$\frac{d}{dx} y \times ( x ^{2} - x + 1 ) = \frac{d}{dx} ( {x}^{4 } + {x}^{2} + 1)$

=> y d/dx ( x² - x + 1 ) + ( x² - x + 1 ) dy/dx = 4x³ + 2x

y ( 2x - 1 ) + ( x ² - x + 1 ) dy/dx = 4x³ + 2x

dy/dx ( x² - x + 1 )= 4x³ + 2x - [ y ( 2x - 1 ) ]

dy/dx ( x² - x + 1 ) = 4x³ + 2x - [ ( x^4 + x² + 1 ) ( 2x - 1 ) / x² - x + 1 ]

dy/dx ( x² - x + 1 ) = 4x³ + 2x - [ 2x^5 + 2x³ + 2x - x^4 - x² - 1 / x² - x + 1 ]

dy/dx ( x² - x + 1 ) = (4x^5 - 4x^4 + 4x³ + 2x³ - 2x² + 2x - 2x^5 - 2x³ - 2x + x^4 + x² + 1 ) / ( x² - x + 1 )

dy/dx ( x² - x + 1 )² = 2x^5 - 3x^4 + 4x³ - x² + 1

$\frac{dy}{dx} = \frac{2 {x}^{5} - 3 {x}^{4} + 4 {x}^{3} - {x}^{2} + 1 }{ {x}^{4} + {x}^{2} + 1 - 2 {x}^{3} - 2x + 2 {x}^{2} }$

${\frac{dy}{dx} = \frac{2 {x}^{5} - 3 {x}^{4} + 4 {x}^{3} - {x}^{2} + 1 }{ {x}^{4} - 2 {x}^{3} + 3{x}^{2} - 2x + 1 }}$

So, dy/dx = 2x + 1

[ After performing Long division, We get it ]

Or

y = (x^4+x^2+1)/(x^2-x+1)

y = ( x^2 + x + 1 )( x^2 - x + 1 ) / x^2 - x + 1

y = x^2 + x + 1

dy/dx = 2x + 1 .


Hope helped!

Answer 2

$\huge\boxed{\fcolorbox{green}{purple}{Solution }}$

y = ( x^4 + x^2 + 1 ) / (x² - x + 1 )

y ( x² - x + 1 ) = x^4 + x² + 1

Now, Differentiating with respect to x

We know that,

d(uv)/dx = v du/dx + u dv/dx

\frac{d}{dx} y \times ( x ^{2} - x + 1 ) = \frac{d}{dx} ( {x}^{4 } + {x}^{2} + 1)

dx

d

y×(x

2

−x+1)=

dx

d

(x

4

+x

2

+1)

=> y d/dx ( x² - x + 1 ) + ( x² - x + 1 ) dy/dx = 4x³ + 2x

y ( 2x - 1 ) + ( x ² - x + 1 ) dy/dx = 4x³ + 2x

dy/dx ( x² - x + 1 )= 4x³ + 2x - [ y ( 2x - 1 ) ]

dy/dx ( x² - x + 1 ) = 4x³ + 2x - [ ( x^4 + x² + 1 ) ( 2x - 1 ) / x² - x + 1 ]

dy/dx ( x² - x + 1 ) = 4x³ + 2x - [ 2x^5 + 2x³ + 2x - x^4 - x² - 1 / x² - x + 1 ]

dy/dx ( x² - x + 1 ) = (4x^5 - 4x^4 + 4x³ + 2x³ - 2x² + 2x - 2x^5 - 2x³ - 2x + x^4 + x² + 1 ) / ( x² - x + 1 )

dy/dx ( x² - x + 1 )² = 2x^5 - 3x^4 + 4x³ - x² + 1

\frac{dy}{dx} = \frac{2 {x}^{5} - 3 {x}^{4} + 4 {x}^{3} - {x}^{2} + 1 }{ {x}^{4} + {x}^{2} + 1 - 2 {x}^{3} - 2x + 2 {x}^{2} }

dx

dy

=

x

4

+x

2

+1−2x

3

−2x+2x

2

2x

5

−3x

4

+4x

3

−x

2

+1

{\frac{dy}{dx} = \frac{2 {x}^{5} - 3 {x}^{4} + 4 {x}^{3} - {x}^{2} + 1 }{ {x}^{4} - 2 {x}^{3} + 3{x}^{2} - 2x + 1 }}

dx

dy

=

x

4

−2x

3

+3x

2

−2x+1

2x

5

−3x

4

+4x

3

−x

2

+1

So, dy/dx = 2x + 1

[ After performing Long division, We get it ]

Or

y = (x^4+x^2+1)/(x^2-x+1)

y = ( x^2 + x + 1 )( x^2 - x + 1 ) / x^2 - x + 1

y = x^2 + x + 1

dy/dx = 2x + 1 .

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