What is the derivative of ln(1/(3x)) ?

The derivative of ln(1/(3x)) is -1/x

What is the first derivative of ln(1/(3x)) ?

The first derivative of ln(1/(3x)) is -1/x

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\frac{d}{d x} (ln (\frac{1}{3 x}))

- \frac{1}{x}

Solution steps

\frac{d}{d x} (ln (\frac{1}{3 x}))

Simplify

ln (\frac{1}{3 x}) : - ln (3 x)

= \frac{d}{d x} (- ln (3 x))

Take the constant out:

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

= - \frac{d}{d x} (ln (3 x))

Apply the chain rule:

\frac{1}{3 x} \frac{d}{d x} (3 x)

= \frac{1}{3 x} \frac{d}{d x} (3 x)

\frac{d}{d x} (3 x) = 3

= - \frac{1}{3 x} \cdot 3

Simplify

- \frac{1}{3 x} \cdot 3 : - \frac{1}{x}

= - \frac{1}{x}

\int sin (3 π) t d t

\frac{d}{d x} (3 e^{- x^{3}})

f (x) = x^{2} - 3 x + 1

\frac{d}{d x} (lo g_{x} (e))

\int_{1}^{x} \frac{1}{t ^{3} + 1} d t

What is the derivative of ln(1/(3x)) ?
The derivative of ln(1/(3x)) is -1/x
What is the first derivative of ln(1/(3x)) ?
The first derivative of ln(1/(3x)) is -1/x