Calculus Examples

$y = x^{2} \ln (8 x)$

Step 1

Find the first derivative.

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Step 1.1

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 $f (x) = x^{2}$ and $g (x) = \ln (8 x)$ .

$f' (x) = x^{2} \frac{d}{d x} (\ln (8 x)) + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.2

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 $g (x) = 8 x$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $8 x$ .

$f' (x) = x^{2} (\frac{d}{d u} (\ln (u)) \frac{d}{d x} (8 x)) + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.2.2

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

$f' (x) = x^{2} (\frac{1}{u} \cdot \frac{d}{d x} (8 x)) + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.2.3

Replace all occurrences of $u$ with $8 x$ .

$f' (x) = x^{2} (\frac{1}{8 x} \cdot \frac{d}{d x} (8 x)) + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.3

Differentiate.

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Step 1.3.1

Combine $\frac{1}{8 x}$ and $x^{2}$ .

$f' (x) = \frac{x^{2}}{8 x} \cdot \frac{d}{d x} (8 x) + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.3.2

Cancel the common factor of $x^{2}$ and $x$ .

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Step 1.3.2.1

Factor $x$ out of $x^{2}$ .

$f' (x) = \frac{x \cdot x}{8 x} \cdot \frac{d}{d x} (8 x) + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.3.2.2

Cancel the common factors.

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Step 1.3.2.2.1

Factor $x$ out of $8 x$ .

$f' (x) = \frac{x \cdot x}{x \cdot 8} \cdot \frac{d}{d x} (8 x) + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.3.2.2.2

Cancel the common factor.

$f' (x) = \frac{x \cdot x}{x \cdot 8} \cdot \frac{d}{d x} (8 x) + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.3.2.2.3

Rewrite the expression.

$f' (x) = \frac{x}{8} \cdot \frac{d}{d x} (8 x) + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.3.3

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$f' (x) = \frac{x}{8} \cdot (8 \frac{d}{d x} (x)) + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.3.4

Simplify terms.

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Step 1.3.4.1

Combine $8$ and $\frac{x}{8}$ .

$f' (x) = \frac{8 x}{8} \cdot \frac{d}{d x} (x) + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.3.4.2

Cancel the common factor of $8$ .

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Step 1.3.4.2.1

Cancel the common factor.

$f' (x) = \frac{8 x}{8} \cdot \frac{d}{d x} (x) + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.3.4.2.2

Divide $x$ by $1$ .

$f' (x) = x \frac{d}{d x} (x) + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.3.5

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

$f' (x) = x \cdot 1 + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.3.6

Multiply $x$ by $1$ .

$f' (x) = x + \ln (8 x) \frac{d}{d x} (x^{2})$

Step 1.3.7

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f' (x) = x + \ln (8 x) (2 x)$

Step 1.3.8

Reorder terms.

$f' (x) = 2 x \ln (8 x) + x$

Step 2

Find the second derivative.

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Step 2.1

By the Sum Rule, the derivative of $2 x \ln (8 x) + x$ with respect to $x$ is $\frac{d}{d x} [2 x \ln (8 x)] + \frac{d}{d x} [x]$ .

$f'' (x) = \frac{d}{d x} (2 x \ln (8 x)) + \frac{d}{d x} (x)$

Step 2.2

Evaluate $\frac{d}{d x} [2 x \ln (8 x)]$ .

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Step 2.2.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x \ln (8 x)$ with respect to $x$ is $2 \frac{d}{d x} [x \ln (8 x)]$ .

$f'' (x) = 2 \frac{d}{d x} (x \ln (8 x)) + \frac{d}{d x} (x)$

Step 2.2.2

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 $g (x) = \ln (8 x)$ .

$f'' (x) = 2 (x \frac{d}{d x} (\ln (8 x)) + \ln (8 x) \frac{d}{d x} (x)) + \frac{d}{d x} (x)$

Step 2.2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 8 x$ .

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Step 2.2.3.1

To apply the Chain Rule, set $u$ as $8 x$ .

$f'' (x) = 2 (x (\frac{d}{d u} (\ln (u)) \frac{d}{d x} (8 x)) + \ln (8 x) \frac{d}{d x} (x)) + \frac{d}{d x} (x)$

Step 2.2.3.2

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

$f'' (x) = 2 (x (\frac{1}{u} \cdot \frac{d}{d x} (8 x)) + \ln (8 x) \frac{d}{d x} (x)) + \frac{d}{d x} (x)$

Step 2.2.3.3

Replace all occurrences of $u$ with $8 x$ .

$f'' (x) = 2 (x (\frac{1}{8 x} \cdot \frac{d}{d x} (8 x)) + \ln (8 x) \frac{d}{d x} (x)) + \frac{d}{d x} (x)$

Step 2.2.4

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$f'' (x) = 2 (x (\frac{1}{8 x} \cdot (8 \frac{d}{d x} (x))) + \ln (8 x) \frac{d}{d x} (x)) + \frac{d}{d x} (x)$

Step 2.2.5

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

$f'' (x) = 2 (x (\frac{1}{8 x} \cdot (8 \cdot 1)) + \ln (8 x) \frac{d}{d x} (x)) + \frac{d}{d x} (x)$

Step 2.2.6

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

$f'' (x) = 2 (x (\frac{1}{8 x} \cdot (8 \cdot 1)) + \ln (8 x) \cdot 1) + \frac{d}{d x} (x)$

Step 2.2.7

Multiply $8$ by $1$ .

$f'' (x) = 2 (x (\frac{1}{8 x} \cdot 8) + \ln (8 x) \cdot 1) + \frac{d}{d x} (x)$

Step 2.2.8

Combine $\frac{1}{8 x}$ and $8$ .

$f'' (x) = 2 (x (\frac{8}{8 x}) + \ln (8 x) \cdot 1) + \frac{d}{d x} (x)$

Step 2.2.9

Cancel the common factor of $8$ .

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Step 2.2.9.1

Cancel the common factor.

$f'' (x) = 2 (x (\frac{8}{8 x}) + \ln (8 x) \cdot 1) + \frac{d}{d x} (x)$

Step 2.2.9.2

Rewrite the expression.

$f'' (x) = 2 (x (\frac{1}{x}) + \ln (8 x) \cdot 1) + \frac{d}{d x} (x)$

Step 2.2.10

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

$f'' (x) = 2 (\frac{x}{x} + \ln (8 x) \cdot 1) + \frac{d}{d x} (x)$

Step 2.2.11

Cancel the common factor of $x$ .

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Step 2.2.11.1

Cancel the common factor.

$f'' (x) = 2 (\frac{x}{x} + \ln (8 x) \cdot 1) + \frac{d}{d x} (x)$

Step 2.2.11.2

Rewrite the expression.

$f'' (x) = 2 (1 + \ln (8 x) \cdot 1) + \frac{d}{d x} (x)$

Step 2.2.12

Multiply $\ln (8 x)$ by $1$ .

$f'' (x) = 2 (1 + \ln (8 x)) + \frac{d}{d x} (x)$

Step 2.3

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

$f'' (x) = 2 (1 + \ln (8 x)) + 1$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$f'' (x) = 2 \cdot 1 + 2 \ln (8 x) + 1$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Multiply $2$ by $1$ .

$f'' (x) = 2 + 2 \ln (8 x) + 1$

Step 2.4.2.2

Add $2$ and $1$ .

$f'' (x) = 2 \ln (8 x) + 3$

Step 3

Find the third derivative.

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Step 3.1

By the Sum Rule, the derivative of $2 \ln (8 x) + 3$ with respect to $x$ is $\frac{d}{d x} [2 \ln (8 x)] + \frac{d}{d x} [3]$ .

$f''' (x) = \frac{d}{d x} (2 \ln (8 x)) + \frac{d}{d x} (3)$

Step 3.2

Evaluate $\frac{d}{d x} [2 \ln (8 x)]$ .

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Step 3.2.1

Since $2$ is constant with respect to $x$ , the derivative of $2 \ln (8 x)$ with respect to $x$ is $2 \frac{d}{d x} [\ln (8 x)]$ .

$f''' (x) = 2 \frac{d}{d x} (\ln (8 x)) + \frac{d}{d x} (3)$

Step 3.2.2

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 $g (x) = 8 x$ .

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Step 3.2.2.1

To apply the Chain Rule, set $u$ as $8 x$ .

$f''' (x) = 2 (\frac{d}{d u} (\ln (u)) \frac{d}{d x} (8 x)) + \frac{d}{d x} (3)$

Step 3.2.2.2

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

$f''' (x) = 2 (\frac{1}{u} \cdot \frac{d}{d x} (8 x)) + \frac{d}{d x} (3)$

Step 3.2.2.3

Replace all occurrences of $u$ with $8 x$ .

$f''' (x) = 2 (\frac{1}{8 x} \cdot \frac{d}{d x} (8 x)) + \frac{d}{d x} (3)$

Step 3.2.3

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$f''' (x) = 2 (\frac{1}{8 x} \cdot (8 \frac{d}{d x} (x))) + \frac{d}{d x} (3)$

Step 3.2.4

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

$f''' (x) = 2 (\frac{1}{8 x} \cdot (8 \cdot 1)) + \frac{d}{d x} (3)$

Step 3.2.5

Multiply $8$ by $1$ .

$f''' (x) = 2 (\frac{1}{8 x} \cdot 8) + \frac{d}{d x} (3)$

Step 3.2.6

Combine $\frac{1}{8 x}$ and $8$ .

$f''' (x) = 2 (\frac{8}{8 x}) + \frac{d}{d x} (3)$

Step 3.2.7

Cancel the common factor of $8$ .

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Step 3.2.7.1

Cancel the common factor.

$f''' (x) = 2 (\frac{8}{8 x}) + \frac{d}{d x} (3)$

Step 3.2.7.2

Rewrite the expression.

$f''' (x) = 2 (\frac{1}{x}) + \frac{d}{d x} (3)$

Step 3.2.8

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

$f''' (x) = \frac{2}{x} + \frac{d}{d x} (3)$

Step 3.3

Differentiate using the Constant Rule.

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Step 3.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

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

Step 3.3.2

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

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

Step 4

Find the fourth derivative.

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Step 4.1

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

$f^{4} (x) = 2 \frac{d}{d x} (\frac{1}{x})$

Step 4.2

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

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

Step 4.3

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

$f^{4} (x) = 2 (- x^{- 2})$

Step 4.4

Multiply $- 1$ by $2$ .

$f^{4} (x) = - 2 x^{- 2}$

Step 4.5

Simplify.

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Step 4.5.1

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

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

Step 4.5.2

Combine terms.

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Step 4.5.2.1

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

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

Step 4.5.2.2

Move the negative in front of the fraction.

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