What is the derivative of (3x^2/(x^2-4)) ?

The derivative of (3x^2/(x^2-4)) is -(24x)/((x^2-4)^2)

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

The first derivative of (3x^2/(x^2-4)) is -(24x)/((x^2-4)^2)

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

- \frac{24 x}{( x ^{2} - 4 ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

3 \cdot \frac{2 x ( x ^{2} - 4 ) - 2 x x ^{2}}{( x ^{2} - 4 ) ^{2}} : - \frac{24 x}{( x ^{2} - 4 ) ^{2}}

= - \frac{24 x}{( x ^{2} - 4 ) ^{2}}

\frac{d y}{d x} = cos (x - y)

n = 0 \sum \infty \frac{n + 6}{6}

\frac{\partial}{\partial x} (e^{x^{2} y^{3}})

\frac{d}{d x} (12 e^{2} x)

\frac{\partial}{\partial x} (y^{2} z cos (x))

What is the derivative of (3x^2/(x^2-4)) ?
The derivative of (3x^2/(x^2-4)) is -(24x)/((x^2-4)^2)
What is the first derivative of (3x^2/(x^2-4)) ?
The first derivative of (3x^2/(x^2-4)) is -(24x)/((x^2-4)^2)