Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

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 $g (x) = 4 x$ .

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Step 1.1.1

To apply the Chain Rule, set $u$ as $4 x$ .

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

Step 1.1.2

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

$\frac{1}{u} \frac{d}{d x} [4 x]$

Step 1.1.3

Replace all occurrences of $u$ with $4 x$ .

$\frac{1}{4 x} \frac{d}{d x} [4 x]$

Step 1.2

Differentiate.

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Step 1.2.1

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

$\frac{1}{4 x} (4 \frac{d}{d x} [x])$

Step 1.2.2

Simplify terms.

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Step 1.2.2.1

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

$\frac{4}{4 x} \frac{d}{d x} [x]$

Step 1.2.2.2

Cancel the common factor of $4$ .

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Step 1.2.2.2.1

Cancel the common factor.

$\frac{4}{4 x} \frac{d}{d x} [x]$

Step 1.2.2.2.2

Rewrite the expression.

$\frac{1}{x} \frac{d}{d x} [x]$

Step 1.2.3

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

$\frac{1}{x} \cdot 1$

Step 1.2.4

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

$\frac{1}{x}$

Step 2

Find the second derivative of the function.

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

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

Step 4

Since there is no value of $x$ that makes the first derivative equal to $0$ , there are no local extrema.

No Local Extrema

Step 5

No Local Extrema

Step 6