What is the derivative of y=((x-1)/(x^2+x+1))^4 ?

The derivative of y=((x-1)/(x^2+x+1))^4 is (4(x-1)^3(-x^2+2x+2))/((x^2+x+1)^5)

What is the first derivative of y=((x-1)/(x^2+x+1))^4 ?

The first derivative of y=((x-1)/(x^2+x+1))^4 is (4(x-1)^3(-x^2+2x+2))/((x^2+x+1)^5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

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

\frac{4 ( x - 1 ) ^{3} ( - x ^{2} + 2 x + 2 )}{( x ^{2} + x + 1 ) ^{5}}

Solution steps

\frac{d}{d x} ((\frac{x - 1}{x ^{2} + x + 1})^{4})

Apply the chain rule:

4 (\frac{x - 1}{x ^{2} + x + 1})^{3} \frac{d}{d x} (\frac{x - 1}{x ^{2} + x + 1})

= 4 (\frac{x - 1}{x ^{2} + x + 1})^{3} \frac{d}{d x} (\frac{x - 1}{x ^{2} + x + 1})

\frac{d}{d x} (\frac{x - 1}{x ^{2} + x + 1}) = \frac{- x ^{2} + 2 x + 2}{( x ^{2} + x + 1 ) ^{2}}

= 4 (\frac{x - 1}{x ^{2} + x + 1})^{3} \frac{- x ^{2} + 2 x + 2}{( x ^{2} + x + 1 ) ^{2}}

Simplify

4 (\frac{x - 1}{x ^{2} + x + 1})^{3} \frac{- x ^{2} + 2 x + 2}{( x ^{2} + x + 1 ) ^{2}} : \frac{4 ( x - 1 ) ^{3} ( - x ^{2} + 2 x + 2 )}{( x ^{2} + x + 1 ) ^{5}}

= \frac{4 ( x - 1 ) ^{3} ( - x ^{2} + 2 x + 2 )}{( x ^{2} + x + 1 ) ^{5}}

y^{^{'}} + 2 x y = 2 x

\int 5 e^{x} 1 + e^{x} d x

x \to - 9 lim (\frac{x ^{2} - 3}{9 - x})

\int_{0}^{\frac{π}{2}} 12 cos^{5} (x) d x

\int \frac{x ^{3}}{3 x ^{2} + 2} d x

What is the derivative of y=((x-1)/(x^2+x+1))^4 ?
The derivative of y=((x-1)/(x^2+x+1))^4 is (4(x-1)^3(-x^2+2x+2))/((x^2+x+1)^5)
What is the first derivative of y=((x-1)/(x^2+x+1))^4 ?
The first derivative of y=((x-1)/(x^2+x+1))^4 is (4(x-1)^3(-x^2+2x+2))/((x^2+x+1)^5)