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

The derivative of ((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 ((x-1/(x^2+x+1))^4) ?

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

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\frac{d}{d x} ((\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}}

x \to - 10 lim (10 [∣ x ∣] - 9)

\int sin (x) + 3 d x

f (x) = x^{2} - x, at x = 3

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

lo g_{b} (x)

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