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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

y = 2 x^{2}, y = 2 x + 6

ln (x)

\frac{d}{d x} (\frac{e ^{x} - e ^{- x}}{5})

\frac{x d y}{d x} = y + x^{2} - y^{2}

\int_{- \frac{1}{4}}^{\frac{1}{4}} \frac{1}{1 - x ^{2}} d x

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