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

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

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

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

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

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

\frac{3 s - 15}{2 s ^{2} - 4 s + 10}

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} - 4 x + 3})

\frac{d}{d x} (cos (x) + sin (x) + x)

\int \frac{x}{25 + x ^{4}} d x

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