What is the derivative of ((y-3))/(y^2-3y+9) ?

The derivative of ((y-3))/(y^2-3y+9) is (-y^2+6y)/((y^2-3y+9)^2)

What is the first derivative of ((y-3))/(y^2-3y+9) ?

The first derivative of ((y-3))/(y^2-3y+9) is (-y^2+6y)/((y^2-3y+9)^2)

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derivative of

\frac{( y - 3 )}{y ^{2} - 3 y + 9}

\frac{- y ^{2} + 6 y}{( y ^{2} - 3 y + 9 ) ^{2}}

Solution steps

\frac{d}{d y} (\frac{y - 3}{y ^{2} - 3 y + 9})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d y} ( y - 3 ) ( y ^{2} - 3 y + 9 ) - \frac{d}{d y} ( y ^{2} - 3 y + 9 ) ( y - 3 )}{( y ^{2} - 3 y + 9 ) ^{2}}

\frac{d}{d y} (y - 3) = 1

\frac{d}{d y} (y^{2} - 3 y + 9) = 2 y - 3

= \frac{1 \cdot ( y ^{2} - 3 y + 9 ) - ( 2 y - 3 ) ( y - 3 )}{( y ^{2} - 3 y + 9 ) ^{2}}

Simplify

\frac{1 \cdot ( y ^{2} - 3 y + 9 ) - ( 2 y - 3 ) ( y - 3 )}{( y ^{2} - 3 y + 9 ) ^{2}} : \frac{- y ^{2} + 6 y}{( y ^{2} - 3 y + 9 ) ^{2}}

= \frac{- y ^{2} + 6 y}{( y ^{2} - 3 y + 9 ) ^{2}}

y^{^{''}} + 4 y = t - 2 sin (2 t)

\frac{d}{d y} (x^{2} y sin^{3} (x y^{2}))

x \to 0 lim (arcsin (e^{x}))

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

\frac{d}{d x} (\frac{5 x ^{4}}{2 - x ^{2}})

What is the derivative of ((y-3))/(y^2-3y+9) ?
The derivative of ((y-3))/(y^2-3y+9) is (-y^2+6y)/((y^2-3y+9)^2)
What is the first derivative of ((y-3))/(y^2-3y+9) ?
The first derivative of ((y-3))/(y^2-3y+9) is (-y^2+6y)/((y^2-3y+9)^2)