Calculus Examples

$f (x) = \frac{\ln (x)}{9 x}$

Step 1

Find the first derivative.

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Since $\frac{1}{9}$ is constant with respect to $x$ , the derivative of $\frac{\ln (x)}{9 x}$ with respect to $x$ is $\frac{1}{9} \frac{d}{d x} [\frac{\ln (x)}{x}]$ .

$f' (x) = \frac{1}{9} \cdot \frac{d}{d x} (\frac{\ln (x)}{x})$

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = \ln (x)$ and $g (x) = x$ .

$f' (x) = \frac{1}{9} \cdot \frac{x \frac{d}{d x} (\ln (x)) - \ln (x) \frac{d}{d x} x}{x^{2}}$

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

$f' (x) = \frac{1}{9} \cdot \frac{x (\frac{1}{x}) - \ln (x) \frac{d}{d x} x}{x^{2}}$

Differentiate using the Power Rule.

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Combine $x$ and $\frac{1}{x}$ .

$f' (x) = \frac{1}{9} \cdot \frac{\frac{x}{x} - \ln (x) \frac{d}{d x} x}{x^{2}}$

Cancel the common factor of $x$ .

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Cancel the common factor.

$f' (x) = \frac{1}{9} \cdot \frac{\frac{x}{x} - \ln (x) \frac{d}{d x} x}{x^{2}}$

Rewrite the expression.

$f' (x) = \frac{1}{9} \cdot \frac{1 - \ln (x) \frac{d}{d x} x}{x^{2}}$

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

$f' (x) = \frac{1}{9} \cdot \frac{1 - \ln (x) \cdot 1}{x^{2}}$

Combine fractions.

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Multiply $- 1$ by $1$ .

$f' (x) = \frac{1}{9} \cdot \frac{1 - \ln (x)}{x^{2}}$

Multiply $\frac{1}{9}$ by $\frac{1 - \ln (x)}{x^{2}}$ .

$f' (x) = \frac{1 - \ln (x)}{9 x^{2}}$

Step 2

Find the second derivative.

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Since $\frac{1}{9}$ is constant with respect to $x$ , the derivative of $\frac{1 - \ln (x)}{9 x^{2}}$ with respect to $x$ is $\frac{1}{9} \frac{d}{d x} [\frac{1 - \ln (x)}{x^{2}}]$ .

$f'' (x) = \frac{1}{9} \cdot \frac{d}{d x} (\frac{1 - \ln (x)}{x^{2}})$

$f'' (x) = \frac{1}{9} \cdot \frac{x^{2} \frac{d}{d x} (1 - \ln (x)) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{{(x^{2})}^{2}}$

Differentiate.

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Multiply the exponents in ${(x^{2})}^{2}$ .

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Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f'' (x) = \frac{1}{9} \cdot \frac{x^{2} \frac{d}{d x} (1 - \ln (x)) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{2 \cdot 2}}$

Multiply $2$ by $2$ .

$f'' (x) = \frac{1}{9} \cdot \frac{x^{2} \frac{d}{d x} (1 - \ln (x)) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

By the Sum Rule, the derivative of $1 - \ln (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- \ln (x)]$ .

$f'' (x) = \frac{1}{9} \cdot \frac{x^{2} (\frac{d}{d x} (1) + \frac{d}{d x} (- \ln (x))) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f'' (x) = \frac{1}{9} \cdot \frac{x^{2} (0 + \frac{d}{d x} (- \ln (x))) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Add $0$ and $\frac{d}{d x} [- \ln (x)]$ .

$f'' (x) = \frac{1}{9} \cdot \frac{x^{2} \frac{d}{d x} (- \ln (x)) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Since $- 1$ is constant with respect to $x$ , the derivative of $- \ln (x)$ with respect to $x$ is $- \frac{d}{d x} [\ln (x)]$ .

$f'' (x) = \frac{1}{9} \cdot \frac{x^{2} (- \frac{d}{d x} \ln (x)) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

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

$f'' (x) = \frac{1}{9} \cdot \frac{x^{2} (- \frac{1}{x}) - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Differentiate using the Power Rule.

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Combine $x^{2}$ and $\frac{1}{x}$ .

$f'' (x) = \frac{1}{9} \cdot \frac{- \frac{x^{2}}{x} - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Cancel the common factor of $x^{2}$ and $x$ .

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Factor $x$ out of $x^{2}$ .

$f'' (x) = \frac{1}{9} \cdot \frac{- \frac{x \cdot x}{x} - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Cancel the common factors.

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Raise $x$ to the power of $1$ .

$f'' (x) = \frac{1}{9} \cdot \frac{- \frac{x \cdot x}{x} - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Factor $x$ out of $x^{1}$ .

$f'' (x) = \frac{1}{9} \cdot \frac{- \frac{x \cdot x}{x \cdot 1} - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Cancel the common factor.

$f'' (x) = \frac{1}{9} \cdot \frac{- \frac{x \cdot x}{x \cdot 1} - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Rewrite the expression.

$f'' (x) = \frac{1}{9} \cdot \frac{- \frac{x}{1} - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

Divide $x$ by $1$ .

$f'' (x) = \frac{1}{9} \cdot \frac{- x - (1 - \ln (x)) \frac{d}{d x} x^{2}}{x^{4}}$

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

$f'' (x) = \frac{1}{9} \cdot \frac{- x - (1 - \ln (x)) (2 x)}{x^{4}}$

Simplify with factoring out.

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Multiply $2$ by $- 1$ .

$f'' (x) = \frac{1}{9} \cdot \frac{- x - 2 (1 - \ln (x)) x}{x^{4}}$

Factor $x$ out of $- x - 2 (1 - \ln (x)) x$ .

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Factor $x$ out of $- x$ .

$f'' (x) = \frac{1}{9} \cdot \frac{x \cdot - 1 - 2 (1 - \ln (x)) x}{x^{4}}$

Factor $x$ out of $- 2 (1 - \ln (x)) x$ .

$f'' (x) = \frac{1}{9} \cdot \frac{x \cdot - 1 + x (- 2 (1 - \ln (x)))}{x^{4}}$

Factor $x$ out of $x \cdot - 1 + x (- 2 (1 - \ln (x)))$ .

$f'' (x) = \frac{1}{9} \cdot \frac{x (- 1 - 2 (1 - \ln (x)))}{x^{4}}$

Cancel the common factors.

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Factor $x$ out of $x^{4}$ .

$f'' (x) = \frac{1}{9} \cdot \frac{x (- 1 - 2 (1 - \ln (x)))}{x \cdot x^{3}}$

Cancel the common factor.

$f'' (x) = \frac{1}{9} \cdot \frac{x (- 1 - 2 (1 - \ln (x)))}{x \cdot x^{3}}$

Rewrite the expression.

$f'' (x) = \frac{1}{9} \cdot \frac{- 1 - 2 (1 - \ln (x))}{x^{3}}$

Multiply $\frac{1}{9}$ by $\frac{- 1 - 2 (1 - \ln (x))}{x^{3}}$ .

$f'' (x) = \frac{- 1 - 2 (1 - \ln (x))}{9 x^{3}}$

Simplify.

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Apply the distributive property.

$f'' (x) = \frac{- 1 - 2 \cdot 1 - 2 (- \ln (x))}{9 x^{3}}$

Simplify the numerator.

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Simplify each term.

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Multiply $- 2$ by $1$ .

$f'' (x) = \frac{- 1 - 2 - 2 (- \ln (x))}{9 x^{3}}$

Multiply $- 2 (- \ln (x))$ .

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Multiply $- 1$ by $- 2$ .

$f'' (x) = \frac{- 1 - 2 + 2 \ln (x)}{9 x^{3}}$

Simplify $2 \ln (x)$ by moving $2$ inside the logarithm.

$f'' (x) = \frac{- 1 - 2 + \ln (x^{2})}{9 x^{3}}$

Subtract $2$ from $- 1$ .

$f'' (x) = \frac{- 3 + \ln (x^{2})}{9 x^{3}}$

Rewrite $- 3$ as $- 1 (3)$ .

$f'' (x) = \frac{- 1 \cdot 3 + \ln (x^{2})}{9 x^{3}}$

Factor $- 1$ out of $\ln (x^{2})$ .

$f'' (x) = \frac{- 1 \cdot 3 - 1 (- \ln (x^{2}))}{9 x^{3}}$

Factor $- 1$ out of $- 1 (3) - 1 (- \ln (x^{2}))$ .

$f'' (x) = \frac{- 1 (3 - \ln (x^{2}))}{9 x^{3}}$

Move the negative in front of the fraction.

$f'' (x) = - \frac{3 - \ln (x^{2})}{9 x^{3}}$

Step 3

The second derivative of $f (x)$ with respect to $x$ is $- \frac{3 - \ln (x^{2})}{9 x^{3}}$ .

$- \frac{3 - \ln (x^{2})}{9 x^{3}}$