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

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

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

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

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

- \frac{1}{x}

Solution steps

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

Simplify

ln (\frac{1}{2 x}) : - ln (2 x)

= \frac{d}{d x} (- ln (2 x))

Take the constant out:

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

= - \frac{d}{d x} (ln (2 x))

Apply the chain rule:

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

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

\frac{d}{d x} (2 x) = 2

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

Simplify

- \frac{1}{2 x} \cdot 2 : - \frac{1}{x}

= - \frac{1}{x}

ln (3^{x})

y = 3 e^{x} cos (x)

\int (\frac{x}{7} - \frac{3}{4} x^{4}) d x

2 t^{2} y^{^{''}} + 3 t y^{^{'}} - y = 0

\int_{0}^{\infty} e^{- \frac{x}{2}} d x

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