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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (x - 1) = 1

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

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

Simplify

\frac{1 \cdot ( x ^{2} - x + 1 ) - ( 2 x - 1 ) ( x - 1 )}{( x ^{2} - x + 1 ) ^{2}} : \frac{- x ^{2} + 2 x}{( x ^{2} - x + 1 ) ^{2}}

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

\int \frac{x}{- x - 1} d x

\int 4 x^{3} + 2 x y^{2} ln (x^{2} + y^{2}) d x

(x^{2} - x y) d x + x d y = 0

\int \frac{x ^{3} + 2}{x ^{2} + 8 x + 7} d x

\int \frac{x + 1}{x - 2} d x

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