Calculus Examples

$y = 2 x - \ln (2 x)$

Step 1

Write $y = 2 x - \ln (2 x)$ as a function.

$f (x) = 2 x - \ln (2 x)$

Step 2

Find the second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

By the Sum Rule, the derivative of $2 x - \ln (2 x)$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- \ln (2 x)]$ .

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

Step 2.1.2

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 2.1.2.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 2.1.2.2

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

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

Step 2.1.2.3

Multiply $2$ by $1$ .

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

Step 2.1.3

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

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Step 2.1.3.1

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

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

Step 2.1.3.2

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 $g (x) = 2 x$ .

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Step 2.1.3.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

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

Step 2.1.3.2.2

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

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

Step 2.1.3.2.3

Replace all occurrences of $u$ with $2 x$ .

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

Step 2.1.3.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 2.1.3.4

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

$2 - (\frac{1}{2 x} (2 \cdot 1))$

Step 2.1.3.5

Multiply $2$ by $1$ .

$2 - (\frac{1}{2 x} \cdot 2)$

Step 2.1.3.6

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

$2 - \frac{2}{2 x}$

Step 2.1.3.7

Cancel the common factor of $2$ .

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Step 2.1.3.7.1

Cancel the common factor.

$2 - \frac{2}{2 x}$

Step 2.1.3.7.2

Rewrite the expression.

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

Step 2.1.4

Reorder terms.

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

Step 2.2

Find the second derivative.

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Step 2.2.1

By the Sum Rule, the derivative of $- \frac{1}{x} + 2$ with respect to $x$ is $\frac{d}{d x} [- \frac{1}{x}] + \frac{d}{d x} [2]$ .

$\frac{d}{d x} [- \frac{1}{x}] + \frac{d}{d x} [2]$

Step 2.2.2

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

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Step 2.2.2.1

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

$- \frac{d}{d x} [\frac{1}{x}] + \frac{1}{x} \frac{d}{d x} [- 1] + \frac{d}{d x} [2]$

Step 2.2.2.2

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

$- \frac{d}{d x} [x^{- 1}] + \frac{1}{x} \frac{d}{d x} [- 1] + \frac{d}{d x} [2]$

Step 2.2.2.3

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

$- - x^{- 2} + \frac{1}{x} \frac{d}{d x} [- 1] + \frac{d}{d x} [2]$

Step 2.2.2.4

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

$- - x^{- 2} + \frac{1}{x} \cdot 0 + \frac{d}{d x} [2]$

Step 2.2.2.5

Multiply $- 1$ by $- 1$ .

$1 x^{- 2} + \frac{1}{x} \cdot 0 + \frac{d}{d x} [2]$

Step 2.2.2.6

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

$x^{- 2} + \frac{1}{x} \cdot 0 + \frac{d}{d x} [2]$

Step 2.2.2.7

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

$x^{- 2} + 0 + \frac{d}{d x} [2]$

Step 2.2.2.8

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

$x^{- 2} + \frac{d}{d x} [2]$

Step 2.2.3

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$x^{- 2} + 0$

Step 2.2.4

Simplify.

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Step 2.2.4.1

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

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

Step 2.2.4.2

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

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

Step 2.3

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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Step 3.1

Set the second derivative equal to $0$ .

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

Step 3.2

Set the numerator equal to zero.

$1 = 0$

Step 3.3

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