Calculus Examples

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

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

$g (x) = 4 - \ln (x)$ .

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Step 1.1.1

To apply the Chain Rule, set

$u$ as

$4 - \ln (x)$ .

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

Step 1.1.2

The derivative of

$\ln (u)$ with respect to

$u$ is

$\frac{1}{u}$ .

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

Step 1.1.3

Replace all occurrences of

$u$ with

$4 - \ln (x)$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

By the Sum Rule, the derivative of

$4 - \ln (x)$ with respect to

$x$ is

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

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

Step 1.2.2

Since

$4$ is constant with respect to

$x$ , the derivative of

$4$ with respect to

$x$ is

$0$ .

$\frac{1}{4 - \ln (x)} (0 + \frac{d}{d x} [- \ln (x)])$

Step 1.2.3

Add

$0$ and

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

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

Step 1.2.4

Since

$- 1$ is constant with respect to

$x$ , the derivative of

$- \ln (x)$ with respect to

$x$ is

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

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

Step 1.3

The derivative of

$\ln (x)$ with respect to

$x$ is

$\frac{1}{x}$ .

$\frac{1}{4 - \ln (x)} (- \frac{1}{x})$

Step 1.4

Multiply

$\frac{1}{4 - \ln (x)}$ by

$\frac{1}{x}$ .

$- \frac{1}{(4 - \ln (x)) x}$

Step 1.5

Simplify.

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Step 1.5.1

Apply the distributive property.

$- \frac{1}{4 x - \ln (x) x}$

Step 1.5.2

Factor

$x$ out of

$4 x - \ln (x) x$ .

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Step 1.5.2.1

Factor

$x$ out of

$4 x$ .

$- \frac{1}{x \cdot 4 - \ln (x) x}$

Step 1.5.2.2

Factor

$x$ out of

$- \ln (x) x$ .

$- \frac{1}{x \cdot 4 + x (- \ln (x))}$

Step 1.5.2.3

Factor

$x$ out of

$x \cdot 4 + x (- \ln (x))$ .

$- \frac{1}{x (4 - \ln (x))}$

Step 2

Find the second derivative of the function.

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Step 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

$g (x) = \frac{1}{x (4 - \ln (x))}$ .

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

Step 2.2

Rewrite

$\frac{1}{x (4 - \ln (x))}$ as

${(x (4 - \ln (x)))}^{- 1}$ .

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

Step 2.3

Differentiate using the chain rule, which states that

$\frac{d}{d x} [f (g (x))]$ is

$f' (g (x)) g' (x)$ where

$f (x) = x^{- 1}$ and

$g (x) = x (4 - \ln (x))$ .

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Step 2.3.1

To apply the Chain Rule, set

$u$ as

$x (4 - \ln (x))$ .

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

Step 2.3.2

Differentiate using the Power Rule which states that

$\frac{d}{d u} [u^{n}]$ is

$n u^{n - 1}$ where

$n = - 1$ .

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

Step 2.3.3

Replace all occurrences of

$u$ with

$x (4 - \ln (x))$ .

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

Step 2.4

Multiply.

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Step 2.4.1

Multiply

$- 1$ by

$- 1$ .

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

Step 2.4.2

Multiply

${(x (4 - \ln (x)))}^{- 2}$ by

$1$ .

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

Step 2.5

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

$f (x) = x$ and

$g (x) = 4 - \ln (x)$ .

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

Step 2.6

Differentiate.

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Step 2.6.1

By the Sum Rule, the derivative of

$4 - \ln (x)$ with respect to

$x$ is

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

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

Step 2.6.2

Since

$4$ is constant with respect to

$x$ , the derivative of

$4$ with respect to

$x$ is

$0$ .

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

Step 2.6.3

Add

$0$ and

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

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

Step 2.6.4

Since

$- 1$ is constant with respect to

$x$ , the derivative of

$- \ln (x)$ with respect to

$x$ is

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

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

Step 2.7

The derivative of

$\ln (x)$ with respect to

$x$ is

$\frac{1}{x}$ .

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

Step 2.8

Differentiate.

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Step 2.8.1

Combine

$x$ and

$\frac{1}{x}$ .

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

Step 2.8.2

Reduce the expression by cancelling the common factors.

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Step 2.8.2.1

Cancel the common factor of

$x$ .

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Step 2.8.2.1.1

Cancel the common factor.

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

Step 2.8.2.1.2

Rewrite the expression.

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

Step 2.8.2.2

Multiply

$- 1$ by

$1$ .

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

Step 2.8.3

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 2.8.4

Simplify the expression.

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Step 2.8.4.1

Multiply

$4 - \ln (x)$ by

$1$ .

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

Step 2.8.4.2

Add

$- 1$ and

$4$ .

$f'' (x) = {(x (4 - \ln (x)))}^{- 2} (3 - \ln (x)) + \frac{1}{x (4 - \ln (x))} \cdot \frac{d}{d x} (- 1)$

Step 2.8.5

Since

$- 1$ is constant with respect to

$x$ , the derivative of

$- 1$ with respect to

$x$ is

$0$ .

$f'' (x) = {(x (4 - \ln (x)))}^{- 2} (3 - \ln (x)) + \frac{1}{x (4 - \ln (x))} \cdot 0$

Step 2.8.6

Simplify the expression.

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Step 2.8.6.1

Multiply

$\frac{1}{x (4 - \ln (x))}$ by

$0$ .

$f'' (x) = {(x (4 - \ln (x)))}^{- 2} (3 - \ln (x)) + 0$

Step 2.8.6.2

Add

${(x (4 - \ln (x)))}^{- 2} (3 - \ln (x))$ and

$0$ .

$f'' (x) = {(x (4 - \ln (x)))}^{- 2} (3 - \ln (x))$

Step 2.9

Simplify.

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Step 2.9.1

Rewrite the expression using the negative exponent rule

$b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = \frac{1}{{(x (4 - \ln (x)))}^{2}} \cdot (3 - \ln (x))$

Step 2.9.2

Apply the product rule to

$x (4 - \ln (x))$ .

$f'' (x) = \frac{1}{x^{2} {(4 - \ln (x))}^{2}} \cdot (3 - \ln (x))$

Step 2.9.3

Reorder the factors of

$\frac{1}{x^{2} {(4 - \ln (x))}^{2}} (3 - \ln (x))$ .

$f'' (x) = (3 - \ln (x)) (\frac{1}{x^{2} {(4 - \ln (x))}^{2}})$

Step 2.9.4

Multiply

$3 - \ln (x)$ by

$\frac{1}{x^{2} {(4 - \ln (x))}^{2}}$ .

$f'' (x) = \frac{3 - \ln (x)}{x^{2} {(4 - \ln (x))}^{2}}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to

$0$ and solve.

$- \frac{1}{x (4 - \ln (x))} = 0$

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