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

The derivative of (x-8/(x^2-4)) is (-x^2+16x-4)/((x^2-4)^2)

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

The first derivative of (x-8/(x^2-4)) is (-x^2+16x-4)/((x^2-4)^2)

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

\frac{- x ^{2} + 16 x - 4}{( x ^{2} - 4 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (x - 8) = 1

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

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

Simplify

\frac{1 \cdot ( x ^{2} - 4 ) - 2 x ( x - 8 )}{( x ^{2} - 4 ) ^{2}} : \frac{- x ^{2} + 16 x - 4}{( x ^{2} - 4 ) ^{2}}

= \frac{- x ^{2} + 16 x - 4}{( x ^{2} - 4 ) ^{2}}

x^{2} + x y - y^{2} = 3, (5, 3)

\frac{1}{s ^{2} - 6 s + 10}

f (x) = 2 cos (x^{3}) - \frac{2}{x}

\frac{d}{d x} (6 x^{5} y)

x^{3} + 1

What is the derivative of (x-8/(x^2-4)) ?
The derivative of (x-8/(x^2-4)) is (-x^2+16x-4)/((x^2-4)^2)
What is the first derivative of (x-8/(x^2-4)) ?
The first derivative of (x-8/(x^2-4)) is (-x^2+16x-4)/((x^2-4)^2)