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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

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

\frac{- u ^{2} + 8}{( u ^{2} + 8 ) ^{2}}

f (x) = e^{x} cosh (x)

x^{2} y^{^{''}} - 11 x y^{^{'}} + 36 y = 0

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

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