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

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

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

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

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

- \frac{1}{x + 1}

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

\int x^{3} x^{2} + 43 d x

ln (x e^{x} + 8)

f (x) = 4 (100)^{7}

\int x^{3} - \frac{1}{2} d x

f (x) = - \frac{1}{( x - 5 ) ^{2}}, at x = - 3

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