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}}

\int \frac{1}{2} d x

\int_{1}^{2} \frac{1}{x - 1} d x

\int \frac{16}{x ^{2} - 1} d x

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

\int \frac{3 x ^{2} + 1}{x ^{2} 4 - x ^{2}} d x

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)