What is the derivative of (2x/(3x^2-12)) ?

The derivative of (2x/(3x^2-12)) is -(2(x^2+4))/(3(x^2-4)^2)

What is the first derivative of (2x/(3x^2-12)) ?

The first derivative of (2x/(3x^2-12)) is -(2(x^2+4))/(3(x^2-4)^2)

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\frac{d}{d x} (\frac{2 x}{3 x ^{2} - 12})

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

Solution steps

\frac{d}{d x} (\frac{2 x}{3 x ^{2} - 12})

Take the constant out:

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

= 2 \frac{d}{d x} (\frac{x}{3 x ^{2} - 12})

Apply the Quotient Rule:

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

= 2 \cdot \frac{\frac{d x}{d x} ( 3 x ^{2} - 12 ) - \frac{d}{d x} ( 3 x ^{2} - 12 ) x}{( 3 x ^{2} - 12 ) ^{2}}

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

\frac{d}{d x} (3 x^{2} - 12) = 6 x

= 2 \cdot \frac{1 \cdot ( 3 x ^{2} - 12 ) - 6 xx}{( 3 x ^{2} - 12 ) ^{2}}

Simplify

2 \cdot \frac{1 \cdot ( 3 x ^{2} - 12 ) - 6 xx}{( 3 x ^{2} - 12 ) ^{2}} : - \frac{2 ( x ^{2} + 4 )}{3 ( x ^{2} - 4 ) ^{2}}

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

(x^{3} + 4) (3 x^{2} - 4 x + 2)

3 t^{2} + 15 t + 19

\frac{d y}{d x} = \frac{x y - y}{y + 1}

x \to \infty lim (\frac{x}{3 ^{2 x + 1}})

\frac{d}{d x} (\frac{x}{arctan ( x )})

What is the derivative of (2x/(3x^2-12)) ?
The derivative of (2x/(3x^2-12)) is -(2(x^2+4))/(3(x^2-4)^2)
What is the first derivative of (2x/(3x^2-12)) ?
The first derivative of (2x/(3x^2-12)) is -(2(x^2+4))/(3(x^2-4)^2)