What is the derivative of (6-2x/(x^2-4)) ?

The derivative of (6-2x/(x^2-4)) is (2x^2-12x+8)/((x^2-4)^2)

What is the first derivative of (6-2x/(x^2-4)) ?

The first derivative of (6-2x/(x^2-4)) is (2x^2-12x+8)/((x^2-4)^2)

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\frac{d}{d x} (\frac{6 - 2 x}{x ^{2} - 4})

\frac{2 x ^{2} - 12 x + 8}{( x ^{2} - 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 - 2 x}{x ^{2} - 4})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 6 - 2 x ) ( x ^{2} - 4 ) - \frac{d}{d x} ( x ^{2} - 4 ) ( 6 - 2 x )}{( x ^{2} - 4 ) ^{2}}

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

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

= \frac{( - 2 ) ( x ^{2} - 4 ) - 2 x ( 6 - 2 x )}{( x ^{2} - 4 ) ^{2}}

Simplify

\frac{( - 2 ) ( x ^{2} - 4 ) - 2 x ( 6 - 2 x )}{( x ^{2} - 4 ) ^{2}} : \frac{2 x ^{2} - 12 x + 8}{( x ^{2} - 4 ) ^{2}}

= \frac{2 x ^{2} - 12 x + 8}{( x ^{2} - 4 ) ^{2}}

\frac{d}{d x} (d - \frac{d}{x ^{2}} - a x - b)

\frac{d}{d x} (\frac{1 + cos ( 2 x )}{2})

\int ln ∣ e^{x y} cos (x) ∣ d y

\int t \cdot ln (t + 1) d t

\frac{d}{d x} (\frac{3}{x + 9})

What is the derivative of (6-2x/(x^2-4)) ?
The derivative of (6-2x/(x^2-4)) is (2x^2-12x+8)/((x^2-4)^2)
What is the first derivative of (6-2x/(x^2-4)) ?
The first derivative of (6-2x/(x^2-4)) is (2x^2-12x+8)/((x^2-4)^2)