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}}

f (x) = 3 x^{\frac{1}{3}} - 5 x^{\frac{4}{3}}

y = e^{x}, y = e^{4 x}, x = 1

\frac{1}{( ( s - 1 ) ^{2} + 4 )}

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

y^{^{''}} - 4 y + 4 y = 0

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)