What is the derivative of y=(x^3+8)/x ?

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

What is the first derivative of y=(x^3+8)/x ?

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

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derivative of

y = \frac{x ^{3} + 8}{x}

\frac{2 x ^{3} - 8}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{3} + 8}{x})

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

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

\frac{d x}{d x} = 1

= \frac{3 x ^{2} x - 1 \cdot ( x ^{3} + 8 )}{x ^{2}}

Simplify

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

= \frac{2 x ^{3} - 8}{x ^{2}}

\int \frac{16}{16 + e ^{x}} d x

f (u) = \frac{1}{u}

n = 1 \sum \infty \frac{( ( - 1 ) ^{n} )}{n}

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

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