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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

\int_{0}^{\frac{π}{4}} sin (4 t) d t

ln (\frac{x - 2}{x + 2})

\int \frac{4 x ^{6} + 3 x ^{4}}{x ^{3}} d x

y^{'} = 0.0018 y (2400 - y)

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

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