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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derivative of

\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}}

y^{^{'}} + y = e^{x} y^{2}

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

f (x) = \frac{x}{x - 7}

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{7 + e ^{x}}

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