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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{2 xx - 1 \cdot ( x ^{2} - 4 )}{x ^{2}} : \frac{x ^{2} + 4}{x ^{2}}

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

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

\int (sec^{2} (x) - 4) d x

x^{- 2}, 4

\int \frac{5}{4 x} d x

\int e^{- x - 1} d x

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