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

The derivative of f(x)=(x(x^2+4))/(x-2) is (2x^3-6x^2-8)/((x-2)^2)

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

The first derivative of f(x)=(x(x^2+4))/(x-2) is (2x^3-6x^2-8)/((x-2)^2)

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derivative of

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

\frac{2 x ^{3} - 6 x ^{2} - 8}{( x - 2 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

= \frac{( 3 x ^{2} + 4 ) ( x - 2 ) - 1 \cdot x ( x ^{2} + 4 )}{( x - 2 ) ^{2}}

Simplify

\frac{( 3 x ^{2} + 4 ) ( x - 2 ) - 1 \cdot x ( x ^{2} + 4 )}{( x - 2 ) ^{2}} : \frac{2 x ^{3} - 6 x ^{2} - 8}{( x - 2 ) ^{2}}

= \frac{2 x ^{3} - 6 x ^{2} - 8}{( x - 2 ) ^{2}}

\int 7 x^{- \frac{8}{9}} d x

\frac{y ^{^{'}}}{x ^{3} + y} = \frac{1}{x}

\int \frac{co tan ( ln ( x ) )}{2 x}

\frac{\partial ^{2}}{\partial y ^{2}} (x^{3} + 3 x y^{2})

x \to 0 - lim (\frac{2}{e ^{x} + 1})

What is the derivative of f(x)=(x(x^2+4))/(x-2) ?
The derivative of f(x)=(x(x^2+4))/(x-2) is (2x^3-6x^2-8)/((x-2)^2)
What is the first derivative of f(x)=(x(x^2+4))/(x-2) ?
The first derivative of f(x)=(x(x^2+4))/(x-2) is (2x^3-6x^2-8)/((x-2)^2)