What is the tangent of ln(1-x) ?

The tangent of ln(1-x) is y=-1/(1-a_{0)}x+ln(1-a_{0})+(a_{0})/(1-a_{0)}

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tangent of ln(1-x)

tangent of

ln (1 - x)

y = - \frac{1}{1 - a _{0}} x + ln (1 - a_{0}) + \frac{a _{0}}{1 - a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, ln (1 - a_{0}))

Find the slope of

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

Compute the slope of

y = ln (1 - x)

x = a_{0} : m = - \frac{1}{1 - a _{0}}

Find the line with slope m=

- \frac{1}{1 - a _{0}}

and passing through

(a_{0}, ln (1 - a_{0})) : y = - \frac{1}{1 - a _{0}} x + ln (1 - a_{0}) + \frac{a _{0}}{1 - a _{0}}

y = - \frac{1}{1 - a _{0}} x + ln (1 - a_{0}) + \frac{a _{0}}{1 - a _{0}}

\frac{d}{d t} (ln (2 t + 1))

\int \frac{1}{3 x + 1} d x

\int 64 d x

\int x^{2} (x^{3} + 8)^{2} d x

n = 1 \sum \infty 2 (0.5)^{n - 1}

What is the tangent of ln(1-x) ?
The tangent of ln(1-x) is y=-1/(1-a_{0)}x+ln(1-a_{0})+(a_{0})/(1-a_{0)}