Calculus Examples

$y = 2 x \ln (5 x^{2})$

Step 1

Find the first derivative.

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Step 1.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x \ln (5 x^{2})$ with respect to $x$ is $2 \frac{d}{d x} [x \ln (5 x^{2})]$ .

$2 \frac{d}{d x} [x \ln (5 x^{2})]$

Step 1.2

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 $g (x) = \ln (5 x^{2})$ .

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

Step 1.3

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 $g (x) = 5 x^{2}$ .

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Step 1.3.1

To apply the Chain Rule, set $u$ as $5 x^{2}$ .

$2 (x (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [5 x^{2}]) + \ln (5 x^{2}) \frac{d}{d x} [x])$

Step 1.3.2

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

$2 (x (\frac{1}{u} \frac{d}{d x} [5 x^{2}]) + \ln (5 x^{2}) \frac{d}{d x} [x])$

Step 1.3.3

Replace all occurrences of $u$ with $5 x^{2}$ .

$2 (x (\frac{1}{5 x^{2}} \frac{d}{d x} [5 x^{2}]) + \ln (5 x^{2}) \frac{d}{d x} [x])$

Step 1.4

Differentiate.

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Step 1.4.1

Combine $\frac{1}{5 x^{2}}$ and $x$ .

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

Step 1.4.2

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

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Step 1.4.2.1

Raise $x$ to the power of $1$ .

$2 (\frac{x^{1}}{5 x^{2}} \frac{d}{d x} [5 x^{2}] + \ln (5 x^{2}) \frac{d}{d x} [x])$

Step 1.4.2.2

Factor $x$ out of $x^{1}$ .

$2 (\frac{x \cdot 1}{5 x^{2}} \frac{d}{d x} [5 x^{2}] + \ln (5 x^{2}) \frac{d}{d x} [x])$

Step 1.4.2.3

Cancel the common factors.

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Step 1.4.2.3.1

Factor $x$ out of $5 x^{2}$ .

$2 (\frac{x \cdot 1}{x (5 x)} \frac{d}{d x} [5 x^{2}] + \ln (5 x^{2}) \frac{d}{d x} [x])$

Step 1.4.2.3.2

Cancel the common factor.

$2 (\frac{x \cdot 1}{x (5 x)} \frac{d}{d x} [5 x^{2}] + \ln (5 x^{2}) \frac{d}{d x} [x])$

Step 1.4.2.3.3

Rewrite the expression.

$2 (\frac{1}{5 x} \frac{d}{d x} [5 x^{2}] + \ln (5 x^{2}) \frac{d}{d x} [x])$

Step 1.4.3

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{2}$ with respect to $x$ is $5 \frac{d}{d x} [x^{2}]$ .

$2 (\frac{1}{5 x} (5 \frac{d}{d x} [x^{2}]) + \ln (5 x^{2}) \frac{d}{d x} [x])$

Step 1.4.4

Simplify terms.

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Step 1.4.4.1

Combine $5$ and $\frac{1}{5 x}$ .

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

Step 1.4.4.2

Cancel the common factor of $5$ .

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Step 1.4.4.2.1

Cancel the common factor.

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

Step 1.4.4.2.2

Rewrite the expression.

$2 (\frac{1}{x} \frac{d}{d x} [x^{2}] + \ln (5 x^{2}) \frac{d}{d x} [x])$

Step 1.4.5

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

$2 (\frac{1}{x} (2 x) + \ln (5 x^{2}) \frac{d}{d x} [x])$

Step 1.4.6

Simplify terms.

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Step 1.4.6.1

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

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

Step 1.4.6.2

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

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

Step 1.4.6.3

Cancel the common factor of $x$ .

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Step 1.4.6.3.1

Cancel the common factor.

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

Step 1.4.6.3.2

Divide $2$ by $1$ .

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

Step 1.4.7

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

$2 (2 + \ln (5 x^{2}) \cdot 1)$

Step 1.4.8

Multiply $\ln (5 x^{2})$ by $1$ .

$2 (2 + \ln (5 x^{2}))$

Step 1.5

Simplify.

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Step 1.5.1

Apply the distributive property.

$2 \cdot 2 + 2 \ln (5 x^{2})$

Step 1.5.2

Multiply $2$ by $2$ .

$4 + 2 \ln (5 x^{2})$

Step 1.5.3

Reorder terms.

$f' (x) = 2 \ln (5 x^{2}) + 4$

Step 2

Find the second derivative.

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Step 2.1

By the Sum Rule, the derivative of $2 \ln (5 x^{2}) + 4$ with respect to $x$ is $\frac{d}{d x} [2 \ln (5 x^{2})] + \frac{d}{d x} [4]$ .

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

Step 2.2

Evaluate $\frac{d}{d x} [2 \ln (5 x^{2})]$ .

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Step 2.2.1

Since $2$ is constant with respect to $x$ , the derivative of $2 \ln (5 x^{2})$ with respect to $x$ is $2 \frac{d}{d x} [\ln (5 x^{2})]$ .

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

Step 2.2.2

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 $g (x) = 5 x^{2}$ .

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Step 2.2.2.1

To apply the Chain Rule, set $u$ as $5 x^{2}$ .

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

Step 2.2.2.2

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

$2 (\frac{1}{u} \frac{d}{d x} [5 x^{2}]) + \frac{d}{d x} [4]$

Step 2.2.2.3

Replace all occurrences of $u$ with $5 x^{2}$ .

$2 (\frac{1}{5 x^{2}} \frac{d}{d x} [5 x^{2}]) + \frac{d}{d x} [4]$

Step 2.2.3

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{2}$ with respect to $x$ is $5 \frac{d}{d x} [x^{2}]$ .

$2 (\frac{1}{5 x^{2}} (5 \frac{d}{d x} [x^{2}])) + \frac{d}{d x} [4]$

Step 2.2.4

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

$2 (\frac{1}{5 x^{2}} (5 (2 x))) + \frac{d}{d x} [4]$

Step 2.2.5

Multiply $2$ by $5$ .

$2 (\frac{1}{5 x^{2}} (10 x)) + \frac{d}{d x} [4]$

Step 2.2.6

Combine $10$ and $\frac{1}{5 x^{2}}$ .

$2 (\frac{10}{5 x^{2}} x) + \frac{d}{d x} [4]$

Step 2.2.7

Combine $\frac{10}{5 x^{2}}$ and $x$ .

$2 \frac{10 x}{5 x^{2}} + \frac{d}{d x} [4]$

Step 2.2.8

Cancel the common factor of $10$ and $5$ .

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Step 2.2.8.1

Factor $5$ out of $10 x$ .

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

Step 2.2.8.2

Cancel the common factors.

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Step 2.2.8.2.1

Factor $5$ out of $5 x^{2}$ .

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

Step 2.2.8.2.2

Cancel the common factor.

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

Step 2.2.8.2.3

Rewrite the expression.

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

Step 2.2.9

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

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Step 2.2.9.1

Factor $x$ out of $2 x$ .

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

Step 2.2.9.2

Cancel the common factors.

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Step 2.2.9.2.1

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

$2 \frac{x \cdot 2}{x \cdot x} + \frac{d}{d x} [4]$

Step 2.2.9.2.2

Cancel the common factor.

$2 \frac{x \cdot 2}{x \cdot x} + \frac{d}{d x} [4]$

Step 2.2.9.2.3

Rewrite the expression.

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

Step 2.2.10

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

$\frac{2 \cdot 2}{x} + \frac{d}{d x} [4]$

Step 2.2.11

Multiply $2$ by $2$ .

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

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

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

Step 2.3.2

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

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

Step 3

Find the third derivative.

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Multiply $- 1$ by $4$ .

$- 4 x^{- 2}$

Step 3.5

Simplify.

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Step 3.5.1

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

$- 4 \frac{1}{x^{2}}$

Step 3.5.2

Combine terms.

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Step 3.5.2.1

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

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

Step 3.5.2.2

Move the negative in front of the fraction.

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