What is the first (ln(1-x))^' ?

The first (ln(1-x))^' is -1/(1-x)

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(ln(1-x))^'

(ln (1 - x))^{^{'}}

- \frac{1}{1 - x}

Solution steps

(ln (1 - x))^{'}

Apply the chain rule:

\frac{1}{1 - x} (1 - x)^{'}

= \frac{1}{1 - x} (1 - x)^{'}

(1 - x)^{'} = - 1

= \frac{1}{1 - x} (- 1)

Simplify

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

\int \frac{3}{35 sin ( x ) - 12 cos ( x )} d x

f (x) = 3 x^{2} e^{x} + e^{x} (x^{3} + 8)

y ln (x) - x^{2} y = x y^{^{'}}

\frac{\partial}{\partial y} (x^{3 y})

x \to 0 lim (a^{\frac{1}{x}})