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

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

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

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

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

- \frac{1}{1 + x}

Solution steps

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

Simplify

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

f (x) = arcsec (x)

y^{^{'}} + 7 x e^{y} = 0

\int_{- 1}^{0} 7 x d x

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

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

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