Calculus Examples

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

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 $g (x) = \frac{1}{\ln (x)}$ .

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

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

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

Step 2

Multiply by the reciprocal of the fraction to divide by $\frac{1}{\ln (x)}$ .

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

Step 3

Simplify the expression.

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Step 3.1

Multiply $\ln (x)$ by $1$ .

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

Step 3.2

Rewrite $\frac{1}{\ln (x)}$ as $\ln^{- 1} (x)$ .

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

Step 4

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

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Step 4.1

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

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

Step 4.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = - 1$ .

$\ln (x) (- {u_{2}}^{- 2} \frac{d}{d x} [\ln (x)])$

Step 4.3

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

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

Step 5

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

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

Step 6

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

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

Step 7

Subtract $2$ from $1$ .

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

Step 8

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

$- \ln^{- 1} (x) \frac{1}{x}$

Step 9

Combine fractions.

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Step 9.1

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

$- \frac{\ln^{- 1} (x)}{x}$

Step 9.2

Move $\ln^{- 1} (x)$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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