Find the Derivative - d/dx natural log of 1-3x

$\ln (1 - 3 x)$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 1 - 3 x$ .

To apply the Chain Rule, set $u$ as $1 - 3 x$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [1 - 3 x]$

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$\frac{1}{u} \frac{d}{d x} [1 - 3 x]$

Replace all occurrences of $u$ with $1 - 3 x$ .

$\frac{1}{1 - 3 x} \frac{d}{d x} [1 - 3 x]$

Differentiate.

By the Sum Rule, the derivative of $1 - 3 x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 3 x]$ .

$\frac{1}{1 - 3 x} (\frac{d}{d x} [1] + \frac{d}{d x} [- 3 x])$

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\frac{1}{1 - 3 x} (0 + \frac{d}{d x} [- 3 x])$

Add $0$ and $\frac{d}{d x} [- 3 x]$ .

$\frac{1}{1 - 3 x} \frac{d}{d x} [- 3 x]$

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x$ with respect to $x$ is $- 3 \frac{d}{d x} [x]$ .

$\frac{1}{1 - 3 x} (- 3 \frac{d}{d x} [x])$

Combine fractions.

Combine $- 3$ and $\frac{1}{1 - 3 x}$ .

$\frac{- 3}{1 - 3 x} \frac{d}{d x} [x]$

Move the negative in front of the fraction.

$- \frac{3}{1 - 3 x} \frac{d}{d x} [x]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$- \frac{3}{1 - 3 x} \cdot 1$

Multiply $- 1$ by $1$ .

$- \frac{3}{1 - 3 x}$

Find the Derivative - d/dx natural log of 1-3x

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