Calculus Examples

$x - \ln (x)$

Step 1

Write $x - \ln (x)$ as a function.

$f (x) = x - \ln (x)$

Step 2

Find the second derivative.

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Find the first derivative.

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Differentiate.

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

$f' (x) = \frac{d}{d x} (x) + \frac{d}{d x} (- \ln (x))$

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

$f' (x) = 1 + \frac{d}{d x} (- \ln (x))$

Evaluate $\frac{d}{d x} [- \ln (x)]$ .

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

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

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

$f' (x) = 1 - \frac{1}{x}$

Reorder terms.

$f' (x) = - \frac{1}{x} + 1$

Find the second derivative.

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

$f'' (x) = \frac{d}{d x} (- \frac{1}{x}) + \frac{d}{d x} (1)$

Evaluate $\frac{d}{d x} [- \frac{1}{x}]$ .

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Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{1}{x}$ .

$f'' (x) = - \frac{d}{d x} \cdot \frac{1}{x} + \frac{1}{x} \cdot \frac{d}{d x} (- 1) + \frac{d}{d x} (1)$

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

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

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

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

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

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x^{- 2}$ by $1$ .

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

Multiply $\frac{1}{x}$ by $0$ .

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

Add $x^{- 2}$ and $0$ .

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

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

$f'' (x) = x^{- 2} + 0$

Simplify.

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Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = \frac{1}{x^{2}} + 0$

Add $\frac{1}{x^{2}}$ and $0$ .

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

The second derivative of $f (x)$ with respect to $x$ is $\frac{1}{x^{2}}$ .

$\frac{1}{x^{2}}$

Step 3

Set the second derivative equal to $0$ then solve the equation $\frac{1}{x^{2}} = 0$ .

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Set the second derivative equal to $0$ .

$\frac{1}{x^{2}} = 0$

Set the numerator equal to zero.

$1 = 0$

Since $1 \neq 0$ , there are no solutions.

No solution

Step 4

No values found that can make the second derivative equal to $0$ .

No Inflection Points