Find the Derivative minus d First, divide dx natural log of two x

$\ln (2 x)$

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 $g (x) = 2 x$ .

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To apply the Chain Rule, set $u$ as $2 x$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [2 x]$

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

$\frac{1}{u} \frac{d}{d x} [2 x]$

Replace all occurrences of $u$ with $2 x$ .

$\frac{1}{2 x} \frac{d}{d x} [2 x]$

Differentiate.

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Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$\frac{1}{2 x} (2 \frac{d}{d x} [x])$

Clarify terms.

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Add $2$ and $\frac{1}{2 x}$ .

$\frac{2}{2 x} \frac{d}{d x} [x]$

The common factor should be canceled of $2$ .

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Cancel the common factor.

$\frac{2}{2 x} \frac{d}{d x} [x]$

Reformulate the expression.

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

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

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

First, multiply $\frac{1}{x}$ by $1$ .

$\frac{1}{x}$

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