What is the derivative of (x-6/(x^2-2x+12)) ?

The derivative of (x-6/(x^2-2x+12)) is (-x^2+12x)/((x^2-2x+12)^2)

What is the first derivative of (x-6/(x^2-2x+12)) ?

The first derivative of (x-6/(x^2-2x+12)) is (-x^2+12x)/((x^2-2x+12)^2)

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (x) = cos^{2} (x)

x \to 0 lim (- 2 sin (x))

\frac{\partial}{\partial y} (x^{4} y + x z - y z^{2})

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

y = 3 x^{3} - 2 x, (1, 1)

What is the derivative of (x-6/(x^2-2x+12)) ?
The derivative of (x-6/(x^2-2x+12)) is (-x^2+12x)/((x^2-2x+12)^2)
What is the first derivative of (x-6/(x^2-2x+12)) ?
The first derivative of (x-6/(x^2-2x+12)) is (-x^2+12x)/((x^2-2x+12)^2)