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

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

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

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

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

\frac{x ^{4} - 12 x ^{2}}{( x ^{2} - 4 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{3 x ^{2} ( x ^{2} - 4 ) - 2 x x ^{3}}{( x ^{2} - 4 ) ^{2}} : \frac{x ^{4} - 12 x ^{2}}{( x ^{2} - 4 ) ^{2}}

= \frac{x ^{4} - 12 x ^{2}}{( x ^{2} - 4 ) ^{2}}

\frac{d}{d x} (\frac{1 + csc ( x ^{6} )}{1 - cot ( x ^{6} )})

y = x^{2} - 7 x + 6, x = 2, x = 6

ln (121 sin^{2} (x))

x \to - 5 lim (\frac{\frac{1}{5} + \frac{1}{x}}{5 + x})

x \to 1 lim (\frac{1}{( 1 - x )})

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