What is the d/(dy)((y-6)/(y^2-2y+12)) ?

The d/(dy)((y-6)/(y^2-2y+12)) is (-y^2+12y)/((y^2-2y+12)^2)

What is the first d/(dy)((y-6)/(y^2-2y+12)) ?

The first d/(dy)((y-6)/(y^2-2y+12)) is (-y^2+12y)/((y^2-2y+12)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d y} (\frac{y - 6}{y ^{2} - 2 y + 12})

\frac{- y ^{2} + 12 y}{( y ^{2} - 2 y + 12 ) ^{2}}

Solution steps

\frac{d}{d y} (\frac{y - 6}{y ^{2} - 2 y + 12})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d y} ( y - 6 ) ( y ^{2} - 2 y + 12 ) - \frac{d}{d y} ( y ^{2} - 2 y + 12 ) ( y - 6 )}{( y ^{2} - 2 y + 12 ) ^{2}}

\frac{d}{d y} (y - 6) = 1

\frac{d}{d y} (y^{2} - 2 y + 12) = 2 y - 2

= \frac{1 \cdot ( y ^{2} - 2 y + 12 ) - ( 2 y - 2 ) ( y - 6 )}{( y ^{2} - 2 y + 12 ) ^{2}}

Simplify

\frac{1 \cdot ( y ^{2} - 2 y + 12 ) - ( 2 y - 2 ) ( y - 6 )}{( y ^{2} - 2 y + 12 ) ^{2}} : \frac{- y ^{2} + 12 y}{( y ^{2} - 2 y + 12 ) ^{2}}

= \frac{- y ^{2} + 12 y}{( y ^{2} - 2 y + 12 ) ^{2}}

\int (e^{3 x} + 1)^{2} d x

y^{^{''}} + 2 y^{^{'}} - 3 = 0

x \to - 5 - lim (\frac{3 x}{2 x + 10})

x \to 0 - lim ((\frac{1}{5})^{\frac{1}{x}})

x \to 1 - 0 lim (\frac{1}{1 - x})

What is the d/(dy)((y-6)/(y^2-2y+12)) ?
The d/(dy)((y-6)/(y^2-2y+12)) is (-y^2+12y)/((y^2-2y+12)^2)
What is the first d/(dy)((y-6)/(y^2-2y+12)) ?
The first d/(dy)((y-6)/(y^2-2y+12)) is (-y^2+12y)/((y^2-2y+12)^2)