What is the d/(dy)((y-1)/(y^2-y+1)) ?

The d/(dy)((y-1)/(y^2-y+1)) is (-y^2+2y)/((y^2-y+1)^2)

What is the first d/(dy)((y-1)/(y^2-y+1)) ?

The first d/(dy)((y-1)/(y^2-y+1)) is (-y^2+2y)/((y^2-y+1)^2)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

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

\int \frac{y + 13}{( y + 1 ) ( y + 3 ) ( y - 2 )} d y

\int \frac{1}{- t ^{2} + 8 t - 12} d t

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

x^{- 2}, 2

What is the d/(dy)((y-1)/(y^2-y+1)) ?
The d/(dy)((y-1)/(y^2-y+1)) is (-y^2+2y)/((y^2-y+1)^2)
What is the first d/(dy)((y-1)/(y^2-y+1)) ?
The first d/(dy)((y-1)/(y^2-y+1)) is (-y^2+2y)/((y^2-y+1)^2)