What is the derivative of (8-3x/(x^2-6x+8)) ?

The derivative of (8-3x/(x^2-6x+8)) is (3x^2-16x+24)/((x^2-6x+8)^2)

What is the first derivative of (8-3x/(x^2-6x+8)) ?

The first derivative of (8-3x/(x^2-6x+8)) is (3x^2-16x+24)/((x^2-6x+8)^2)

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\frac{d}{d x} (\frac{8 - 3 x}{x ^{2} - 6 x + 8})

\frac{3 x ^{2} - 16 x + 24}{( x ^{2} - 6 x + 8 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{8 - 3 x}{x ^{2} - 6 x + 8})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 8 - 3 x ) ( x ^{2} - 6 x + 8 ) - \frac{d}{d x} ( x ^{2} - 6 x + 8 ) ( 8 - 3 x )}{( x ^{2} - 6 x + 8 ) ^{2}}

\frac{d}{d x} (8 - 3 x) = - 3

\frac{d}{d x} (x^{2} - 6 x + 8) = 2 x - 6

= \frac{( - 3 ) ( x ^{2} - 6 x + 8 ) - ( 2 x - 6 ) ( 8 - 3 x )}{( x ^{2} - 6 x + 8 ) ^{2}}

Simplify

\frac{( - 3 ) ( x ^{2} - 6 x + 8 ) - ( 2 x - 6 ) ( 8 - 3 x )}{( x ^{2} - 6 x + 8 ) ^{2}} : \frac{3 x ^{2} - 16 x + 24}{( x ^{2} - 6 x + 8 ) ^{2}}

= \frac{3 x ^{2} - 16 x + 24}{( x ^{2} - 6 x + 8 ) ^{2}}

x \to 0 lim (\frac{- 7 + x}{x ^{2}})

x \to 2 lim (3)

\int_{0}^{1} 2 π (x + 1) x^{2} d x

t e^{2 t} cos (3 t)

9 y^{^{'}} - 9 tan (x) y = \frac{20}{cos ( x )}

What is the derivative of (8-3x/(x^2-6x+8)) ?
The derivative of (8-3x/(x^2-6x+8)) is (3x^2-16x+24)/((x^2-6x+8)^2)
What is the first derivative of (8-3x/(x^2-6x+8)) ?
The first derivative of (8-3x/(x^2-6x+8)) is (3x^2-16x+24)/((x^2-6x+8)^2)