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

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

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

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

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derivative of

f (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}{d x} (x^{\frac{1}{10}})

\int u^{2} cos (u) d u

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

\int x^{4} ln (9 x) d x

y^{^{''}} + 4 y^{^{'}} - 3 y = x

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