Question

d/dx [㏑(1-x)] =?

.......................................................................................................

Answer 1

Hey dear ☺️


Here your answer

⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️


d/dx [ln(1-x)]


Here we apply this formula

d/dx(ln x) = 1/x


$\frac{d}{dx} ln(1 - x) \\ \\ = \frac{1}{1 - x} \times \frac{d}{dx}(1 - x) \\ \\ \\ = \frac{1}{1 - x} \times ( - x) \\ \\ \\ = \frac{ - x}{1 - x} \\ \\ \\ \\ \\ hope \: it \: helpful$

Answer 2

$\textbf{Answer}$

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


$\textbf{Topic}$

Calculs -> differentiating logarithmic base questions


$\textbf{Step-by-step solution}$



$\textbf{Explanation}$


The simplest one is using logarithmic derivative


$\frac{d}{dx} ln(x) = \frac{1}{x}$


$\textbf{Using,Chain Rule}$


Let

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


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

$\frac{d}{dx} y = \frac{1}{1 - x} \frac{d}{dx}(1 - x)$

$\frac{d}{dx} y=( \frac{1}{1 - x} )( - 1)$

because

$\frac{d}{dx}( - x ) = - 1$


$\frac{d}{dx} y = - \frac{1}{1 - x}$


$\frac{d}{dx} y = \frac{1}{x - 1}$