What is the derivative of (x^2-1/(2x)) ?

The derivative of (x^2-1/(2x)) is (x^2+1)/(2x^2)

What is the first derivative of (x^2-1/(2x)) ?

The first derivative of (x^2-1/(2x)) is (x^2+1)/(2x^2)

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

\frac{x ^{2} + 1}{2 x ^{2}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{x ^{2} + 1}{2 x ^{2}}

x \to - 2 lim (\frac{2 - x ^{2}}{4 x})

(x^{2} + 1) (\frac{d y}{d x}) + 3 x (y - 1) = 0, y (0) = 6

\frac{1}{3 ( 4 - x ^{3} ) ^{4}}

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

\int_{1}^{e} ln (16 x) d x

What is the derivative of (x^2-1/(2x)) ?
The derivative of (x^2-1/(2x)) is (x^2+1)/(2x^2)
What is the first derivative of (x^2-1/(2x)) ?
The first derivative of (x^2-1/(2x)) is (x^2+1)/(2x^2)