Calculus Examples

$y = \ln (\ln (9 x))$

Step 1

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 $g (x) = \ln (9 x)$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $\ln (9 x)$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [\ln (9 x)]$

Step 1.2

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

$\frac{1}{u_{1}} \frac{d}{d x} [\ln (9 x)]$

Step 1.3

Replace all occurrences of $u_{1}$ with $\ln (9 x)$ .

$\frac{1}{\ln (9 x)} \frac{d}{d x} [\ln (9 x)]$

Step 2

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 $g (x) = 9 x$ .

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Step 2.1

To apply the Chain Rule, set $u_{2}$ as $9 x$ .

$\frac{1}{\ln (9 x)} (\frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d x} [9 x])$

Step 2.2

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

$\frac{1}{\ln (9 x)} (\frac{1}{u_{2}} \frac{d}{d x} [9 x])$

Step 2.3

Replace all occurrences of $u_{2}$ with $9 x$ .

$\frac{1}{\ln (9 x)} (\frac{1}{9 x} \frac{d}{d x} [9 x])$

Step 3

Differentiate.

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Step 3.1

Multiply $\frac{1}{9 x}$ by $\frac{1}{\ln (9 x)}$ .

$\frac{1}{9 x \ln (9 x)} \frac{d}{d x} [9 x]$

Step 3.2

Since $9$ is constant with respect to $x$ , the derivative of $9 x$ with respect to $x$ is $9 \frac{d}{d x} [x]$ .

$\frac{1}{9 x \ln (9 x)} (9 \frac{d}{d x} [x])$

Step 3.3

Simplify terms.

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Step 3.3.1

Combine $9$ and $\frac{1}{9 x \ln (9 x)}$ .

$\frac{9}{9 x \ln (9 x)} \frac{d}{d x} [x]$

Step 3.3.2

Cancel the common factor of $9$ .

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Step 3.3.2.1

Cancel the common factor.

$\frac{9}{9 x \ln (9 x)} \frac{d}{d x} [x]$

Step 3.3.2.2

Rewrite the expression.

$\frac{1}{x \ln (9 x)} \frac{d}{d x} [x]$

Step 3.4

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

$\frac{1}{x \ln (9 x)} \cdot 1$

Step 3.5

Multiply $\frac{1}{x \ln (9 x)}$ by $1$ .

$\frac{1}{x \ln (9 x)}$