Algebra Examples

$\ln (| \ln (5 x) |)$

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 $g (x) = | \ln (5 x) |$ .

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To apply the Chain Rule, set $u_{1}$ as $| \ln (5 x) |$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [| \ln (5 x) |]$

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

$\frac{1}{u_{1}} \frac{d}{d x} [| \ln (5 x) |]$

Replace all occurrences of $u_{1}$ with $| \ln (5 x) |$ .

$\frac{1}{| \ln (5 x) |} \frac{d}{d x} [| \ln (5 x) |]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = | x |$ and $g (x) = \ln (5 x)$ .

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To apply the Chain Rule, set $u_{2}$ as $\ln (5 x)$ .

$\frac{1}{| \ln (5 x) |} (\frac{d}{d u_{2}} [| u_{2} |] \frac{d}{d x} [\ln (5 x)])$

The derivative of $| u_{2} |$ with respect to $u_{2}$ is $\frac{u_{2}}{| u_{2} |}$ .

$\frac{1}{| \ln (5 x) |} (\frac{u_{2}}{| u_{2} |} \frac{d}{d x} [\ln (5 x)])$

Replace all occurrences of $u_{2}$ with $\ln (5 x)$ .

$\frac{1}{| \ln (5 x) |} (\frac{\ln (5 x)}{| \ln (5 x) |} \frac{d}{d x} [\ln (5 x)])$

Step 3

Multiply $\frac{\ln (5 x)}{| \ln (5 x) |}$ by $\frac{1}{| \ln (5 x) |}$ .

$\frac{\ln (5 x)}{| \ln (5 x) | | \ln (5 x) |} \frac{d}{d x} [\ln (5 x)]$

Step 4

To multiply absolute values, multiply the terms inside each absolute value.

$\frac{\ln (5 x)}{| \ln (5 x) \ln (5 x) |} \frac{d}{d x} [\ln (5 x)]$

Step 5

Raise $\ln (5 x)$ to the power of $1$ .

$\frac{\ln (5 x)}{| \ln^{1} (5 x) \ln (5 x) |} \frac{d}{d x} [\ln (5 x)]$

Step 6

Raise $\ln (5 x)$ to the power of $1$ .

$\frac{\ln (5 x)}{| \ln^{1} (5 x) \ln^{1} (5 x) |} \frac{d}{d x} [\ln (5 x)]$

Step 7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{\ln (5 x)}{| {\ln (5 x)}^{1 + 1} |} \frac{d}{d x} [\ln (5 x)]$

Step 8

Add $1$ and $1$ .

$\frac{\ln (5 x)}{| \ln^{2} (5 x) |} \frac{d}{d x} [\ln (5 x)]$

Step 9

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 $g (x) = 5 x$ .

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To apply the Chain Rule, set $u_{3}$ as $5 x$ .

$\frac{\ln (5 x)}{| \ln^{2} (5 x) |} (\frac{d}{d u_{3}} [\ln (u_{3})] \frac{d}{d x} [5 x])$

The derivative of $\ln (u_{3})$ with respect to $u_{3}$ is $\frac{1}{u_{3}}$ .

$\frac{\ln (5 x)}{| \ln^{2} (5 x) |} (\frac{1}{u_{3}} \frac{d}{d x} [5 x])$

Replace all occurrences of $u_{3}$ with $5 x$ .

$\frac{\ln (5 x)}{| \ln^{2} (5 x) |} (\frac{1}{5 x} \frac{d}{d x} [5 x])$

Step 10

Differentiate.

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Multiply $\frac{1}{5 x}$ by $\frac{\ln (5 x)}{| \ln^{2} (5 x) |}$ .

$\frac{\ln (5 x)}{5 x | \ln^{2} (5 x) |} \frac{d}{d x} [5 x]$

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$\frac{\ln (5 x)}{5 x | \ln^{2} (5 x) |} (5 \frac{d}{d x} [x])$

Simplify terms.

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Combine $5$ and $\frac{\ln (5 x)}{5 x | \ln^{2} (5 x) |}$ .

$\frac{5 \ln (5 x)}{5 x | \ln^{2} (5 x) |} \frac{d}{d x} [x]$

Cancel the common factor of $5$ .

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Cancel the common factor.

$\frac{5 \ln (5 x)}{5 x | \ln^{2} (5 x) |} \frac{d}{d x} [x]$

Rewrite the expression.

$\frac{\ln (5 x)}{x | \ln^{2} (5 x) |} \frac{d}{d x} [x]$

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

$\frac{\ln (5 x)}{x | \ln^{2} (5 x) |} \cdot 1$

Multiply $\frac{\ln (5 x)}{x | \ln^{2} (5 x) |}$ by $1$ .

$\frac{\ln (5 x)}{x | \ln^{2} (5 x) |}$

Step 11

Simplify.

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Remove non-negative terms from the absolute value.

$\frac{\ln (5 x)}{x \ln^{2} (5 x)}$

Cancel the common factor of $\ln (5 x)$ and $\ln^{2} (5 x)$ .

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Multiply by $1$ .

$\frac{\ln (5 x) \cdot 1}{x \ln^{2} (5 x)}$

Cancel the common factors.

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Factor $\ln (5 x)$ out of $x \ln^{2} (5 x)$ .

$\frac{\ln (5 x) \cdot 1}{\ln (5 x) (x \ln (5 x))}$

Cancel the common factor.

$\frac{\ln (5 x) \cdot 1}{\ln (5 x) (x \ln (5 x))}$

Rewrite the expression.

$\frac{1}{x \ln (5 x)}$