What is the derivative of (x^3-8/(x^2-4)) ?

The derivative of (x^3-8/(x^2-4)) is (x(x+4))/((x+2)^2)

What is the first derivative of (x^3-8/(x^2-4)) ?

The first derivative of (x^3-8/(x^2-4)) is (x(x+4))/((x+2)^2)

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\frac{d}{d x} (\frac{x ^{3} - 8}{x ^{2} - 4})

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

Solution steps

\frac{d}{d x} (\frac{x ^{3} - 8}{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} - 8 ) ( x ^{2} - 4 ) - \frac{d}{d x} ( x ^{2} - 4 ) ( x ^{3} - 8 )}{( x ^{2} - 4 ) ^{2}}

\frac{d}{d x} (x^{3} - 8) = 3 x^{2}

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

= \frac{3 x ^{2} ( x ^{2} - 4 ) - 2 x ( x ^{3} - 8 )}{( x ^{2} - 4 ) ^{2}}

Simplify

\frac{3 x ^{2} ( x ^{2} - 4 ) - 2 x ( x ^{3} - 8 )}{( x ^{2} - 4 ) ^{2}} : \frac{x ( x + 4 )}{( x + 2 ) ^{2}}

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

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} + 1}{x - 3})

\frac{d}{d x} ((5 + x^{4})^{12})

\int \frac{1}{v ^{2} + 1} d v

\frac{2 s + 1}{3 s ^{2} + 5 s}

What is the derivative of (x^3-8/(x^2-4)) ?
The derivative of (x^3-8/(x^2-4)) is (x(x+4))/((x+2)^2)
What is the first derivative of (x^3-8/(x^2-4)) ?
The first derivative of (x^3-8/(x^2-4)) is (x(x+4))/((x+2)^2)