Calculus Examples

$5 x - 4 \ln (x)$

Step 1

Write $5 x - 4 \ln (x)$ as a function.

$f (x) = 5 x - 4 \ln (x)$

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 $5 x - 4 \ln (x)$ with respect to $x$ is $\frac{d}{d x} [5 x] + \frac{d}{d x} [- 4 \ln (x)]$ .

$\frac{d}{d x} [5 x] + \frac{d}{d x} [- 4 \ln (x)]$

Step 2.1.2

Evaluate $\frac{d}{d x} [5 x]$ .

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Step 2.1.2.1

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

$5 \frac{d}{d x} [x] + \frac{d}{d x} [- 4 \ln (x)]$

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 $n = 1$ .

$5 \cdot 1 + \frac{d}{d x} [- 4 \ln (x)]$

Step 2.1.2.3

Multiply $5$ by $1$ .

$5 + \frac{d}{d x} [- 4 \ln (x)]$

Step 2.1.3

Evaluate $\frac{d}{d x} [- 4 \ln (x)]$ .

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Step 2.1.3.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 \ln (x)$ with respect to $x$ is $- 4 \frac{d}{d x} [\ln (x)]$ .

$5 - 4 \frac{d}{d x} [\ln (x)]$

Step 2.1.3.2

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

$5 - 4 \frac{1}{x}$

Step 2.1.3.3

Combine $- 4$ and $\frac{1}{x}$ .

$5 + \frac{- 4}{x}$

Step 2.1.3.4

Move the negative in front of the fraction.

$5 - \frac{4}{x}$

Step 2.1.4

Reorder terms.

$f' (x) = - \frac{4}{x} + 5$

Step 2.2

Find the second derivative.

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Step 2.2.1

By the Sum Rule, the derivative of $- \frac{4}{x} + 5$ with respect to $x$ is $\frac{d}{d x} [- \frac{4}{x}] + \frac{d}{d x} [5]$ .

$\frac{d}{d x} [- \frac{4}{x}] + \frac{d}{d x} [5]$

Step 2.2.2

Evaluate $\frac{d}{d x} [- \frac{4}{x}]$ .

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Step 2.2.2.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- \frac{4}{x}$ with respect to $x$ is $- 4 \frac{d}{d x} [\frac{1}{x}]$ .

$- 4 \frac{d}{d x} [\frac{1}{x}] + \frac{d}{d x} [5]$

Step 2.2.2.2

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

$- 4 \frac{d}{d x} [x^{- 1}] + \frac{d}{d x} [5]$

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 $n = - 1$ .

$- 4 (- x^{- 2}) + \frac{d}{d x} [5]$

Step 2.2.2.4

Multiply $- 1$ by $- 4$ .

$4 x^{- 2} + \frac{d}{d x} [5]$

Step 2.2.3

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$4 x^{- 2} + 0$

Step 2.2.4

Simplify.

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Step 2.2.4.1

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

$4 \frac{1}{x^{2}} + 0$

Step 2.2.4.2

Combine terms.

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Step 2.2.4.2.1

Combine $4$ and $\frac{1}{x^{2}}$ .

$\frac{4}{x^{2}} + 0$

Step 2.2.4.2.2

Add $\frac{4}{x^{2}}$ and $0$ .

$f'' (x) = \frac{4}{x^{2}}$

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $\frac{4}{x^{2}}$ .

$\frac{4}{x^{2}}$

Step 3

Set the second derivative equal to $0$ then solve the equation $\frac{4}{x^{2}} = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

$\frac{4}{x^{2}} = 0$

Step 3.2

Set the numerator equal to zero.

$4 = 0$

Step 3.3

Since $4 \neq 0$ , there are no solutions.

No solution

Step 4

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

No Inflection Points