What is the derivative of (2x/(x-3)) ?

The derivative of (2x/(x-3)) is -6/((x-3)^2)

What is the first derivative of (2x/(x-3)) ?

The first derivative of (2x/(x-3)) is -6/((x-3)^2)

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\frac{d}{d x} (\frac{2 x}{x - 3})

- \frac{6}{( x - 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x}{x - 3})

Take the constant out:

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

= 2 \frac{d}{d x} (\frac{x}{x - 3})

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

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

\frac{d}{d x} (x - 3) = 1

= 2 \cdot \frac{1 \cdot ( x - 3 ) - 1 \cdot x}{( x - 3 ) ^{2}}

Simplify

2 \cdot \frac{1 \cdot ( x - 3 ) - 1 \cdot x}{( x - 3 ) ^{2}} : - \frac{6}{( x - 3 ) ^{2}}

= - \frac{6}{( x - 3 ) ^{2}}

f (x) = \frac{3 + x}{1 - 3 x}

\int \frac{1}{x ^{4} - 1} d x

x \to 2 lim (\frac{2 - x}{4 - x ^{2}})

\frac{d}{d x} (x^{3} + 2 x y + y^{2} + z^{2} - 5 x - 4 z)

x \to 9 lim (\frac{81}{9 - x})

What is the derivative of (2x/(x-3)) ?
The derivative of (2x/(x-3)) is -6/((x-3)^2)
What is the first derivative of (2x/(x-3)) ?
The first derivative of (2x/(x-3)) is -6/((x-3)^2)