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

The derivative of (x-3/(x^2-3x+2)) is (-x^2+6x-7)/((x^2-3x+2)^2)

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

The first derivative of (x-3/(x^2-3x+2)) is (-x^2+6x-7)/((x^2-3x+2)^2)

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

\frac{- x ^{2} + 6 x - 7}{( x ^{2} - 3 x + 2 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{1 \cdot ( x ^{2} - 3 x + 2 ) - ( 2 x - 3 ) ( x - 3 )}{( x ^{2} - 3 x + 2 ) ^{2}} : \frac{- x ^{2} + 6 x - 7}{( x ^{2} - 3 x + 2 ) ^{2}}

= \frac{- x ^{2} + 6 x - 7}{( x ^{2} - 3 x + 2 ) ^{2}}

\int 4 y^{4} d y

x = y (2 - y), x = 0

\frac{d}{d x} ((6 - x)^{\frac{1}{3}})

\int \frac{3}{x} - \frac{4}{x ^{5}} d x

\frac{9 s}{5 ( s ^{2} + 5 s + 11 )}

What is the derivative of (x-3/(x^2-3x+2)) ?
The derivative of (x-3/(x^2-3x+2)) is (-x^2+6x-7)/((x^2-3x+2)^2)
What is the first derivative of (x-3/(x^2-3x+2)) ?
The first derivative of (x-3/(x^2-3x+2)) is (-x^2+6x-7)/((x^2-3x+2)^2)