What is the derivative of ln(1/(6+x)) ?

The derivative of ln(1/(6+x)) is -1/(6+x)

What is the first derivative of ln(1/(6+x)) ?

The first derivative of ln(1/(6+x)) is -1/(6+x)

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\frac{d}{d x} (ln (\frac{1}{6 + x}))

- \frac{1}{6 + x}

Solution steps

\frac{d}{d x} (ln (\frac{1}{6 + x}))

Simplify

ln (\frac{1}{6 + x}) : - ln (6 + x)

= \frac{d}{d x} (- ln (6 + x))

Take the constant out:

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

= - \frac{d}{d x} (ln (6 + x))

Apply the chain rule:

\frac{1}{6 + x} \frac{d}{d x} (6 + x)

= \frac{1}{6 + x} \frac{d}{d x} (6 + x)

\frac{d}{d x} (6 + x) = 1

= - \frac{1}{6 + x} \cdot 1

Simplify

= - \frac{1}{6 + x}

x \to 2 lim (x + 1)

\int \frac{x ^{2}}{3 x ^{2} + 4} d x

\int 4 x^{3} + 9 sin (x) d x

x \to 3 lim (x + 1)

\int sech^{2} (x) d x

What is the derivative of ln(1/(6+x)) ?
The derivative of ln(1/(6+x)) is -1/(6+x)
What is the first derivative of ln(1/(6+x)) ?
The first derivative of ln(1/(6+x)) is -1/(6+x)