What is the derivative of (x-4/(x^2-2x+8)) ?

The derivative of (x-4/(x^2-2x+8)) is (-x^2+8x)/((x^2-2x+8)^2)

What is the first derivative of (x-4/(x^2-2x+8)) ?

The first derivative of (x-4/(x^2-2x+8)) is (-x^2+8x)/((x^2-2x+8)^2)

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\frac{d}{d x} (\frac{x - 4}{x ^{2} - 2 x + 8})

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

Solution steps

\frac{d}{d x} (\frac{x - 4}{x ^{2} - 2 x + 8})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x - 4 ) ( x ^{2} - 2 x + 8 ) - \frac{d}{d x} ( x ^{2} - 2 x + 8 ) ( x - 4 )}{( x ^{2} - 2 x + 8 ) ^{2}}

\frac{d}{d x} (x - 4) = 1

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

= \frac{1 \cdot ( x ^{2} - 2 x + 8 ) - ( 2 x - 2 ) ( x - 4 )}{( x ^{2} - 2 x + 8 ) ^{2}}

Simplify

\frac{1 \cdot ( x ^{2} - 2 x + 8 ) - ( 2 x - 2 ) ( x - 4 )}{( x ^{2} - 2 x + 8 ) ^{2}} : \frac{- x ^{2} + 8 x}{( x ^{2} - 2 x + 8 ) ^{2}}

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

\frac{\partial}{\partial x} ((x^{2} - y^{2}) ln (x y))

\frac{d}{d x} (9 sin^{2} (x))

y = x^{2} - 14, y = 13 x

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

\frac{10}{( s ( s + 1 ) )}

What is the derivative of (x-4/(x^2-2x+8)) ?
The derivative of (x-4/(x^2-2x+8)) is (-x^2+8x)/((x^2-2x+8)^2)
What is the first derivative of (x-4/(x^2-2x+8)) ?
The first derivative of (x-4/(x^2-2x+8)) is (-x^2+8x)/((x^2-2x+8)^2)