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

$\ln (\ln (6 x))$

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 $g (x) = \ln (6 x)$ .

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To apply the Chain Rule, set $u_{1}$ as $\ln (6 x)$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [\ln (6 x)]$

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

$\frac{1}{u_{1}} \frac{d}{d x} [\ln (6 x)]$

Replace all occurrences of $u_{1}$ with $\ln (6 x)$ .

$\frac{1}{\ln (6 x)} \frac{d}{d x} [\ln (6 x)]$

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 $g (x) = 6 x$ .

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To apply the Chain Rule, set $u_{2}$ as $6 x$ .

$\frac{1}{\ln (6 x)} (\frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d x} [6 x])$

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$\frac{1}{\ln (6 x)} (\frac{1}{u_{2}} \frac{d}{d x} [6 x])$

Replace all occurrences of $u_{2}$ with $6 x$ .

$\frac{1}{\ln (6 x)} (\frac{1}{6 x} \frac{d}{d x} [6 x])$

Differentiate.

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First, multiply $\frac{1}{6 x}$ and $\frac{1}{\ln (6 x)}$ .

$\frac{1}{6 x \ln (6 x)} \frac{d}{d x} [6 x]$

Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$\frac{1}{6 x \ln (6 x)} (6 \frac{d}{d x} [x])$

Clarify terms.

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Add $6$ and $\frac{1}{6 x \ln (6 x)}$ .

$\frac{6}{6 x \ln (6 x)} \frac{d}{d x} [x]$

The common factor should be canceled of $6$ .

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Cancel the common factor.

$\frac{6}{6 x \ln (6 x)} \frac{d}{d x} [x]$

Reformulate the expression.

$\frac{1}{x \ln (6 x)} \frac{d}{d x} [x]$

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

$\frac{1}{x \ln (6 x)} \cdot 1$

Multiply $\frac{1}{x \ln (6 x)}$ by $1$ .

$\frac{1}{x \ln (6 x)}$

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