Calculus Examples

$f (x) = x^{2} \ln (3 x) + 6$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

By the Sum Rule, the derivative of $x^{2} \ln (3 x) + 6$ with respect to $x$ is $\frac{d}{d x} [x^{2} \ln (3 x)] + \frac{d}{d x} [6]$ .

$\frac{d}{d x} [x^{2} \ln (3 x)] + \frac{d}{d x} [6]$

Step 1.1.2

Evaluate $\frac{d}{d x} [x^{2} \ln (3 x)]$ .

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Step 1.1.2.1

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 $f (x) = x^{2}$ and $g (x) = \ln (3 x)$ .

$x^{2} \frac{d}{d x} [\ln (3 x)] + \ln (3 x) \frac{d}{d x} [x^{2}] + \frac{d}{d x} [6]$

Step 1.1.2.2

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 $g (x) = 3 x$ .

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Step 1.1.2.2.1

To apply the Chain Rule, set $u$ as $3 x$ .

$x^{2} (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [3 x]) + \ln (3 x) \frac{d}{d x} [x^{2}] + \frac{d}{d x} [6]$

Step 1.1.2.2.2

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

$x^{2} (\frac{1}{u} \frac{d}{d x} [3 x]) + \ln (3 x) \frac{d}{d x} [x^{2}] + \frac{d}{d x} [6]$

Step 1.1.2.2.3

Replace all occurrences of $u$ with $3 x$ .

$x^{2} (\frac{1}{3 x} \frac{d}{d x} [3 x]) + \ln (3 x) \frac{d}{d x} [x^{2}] + \frac{d}{d x} [6]$

Step 1.1.2.3

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$x^{2} (\frac{1}{3 x} (3 \frac{d}{d x} [x])) + \ln (3 x) \frac{d}{d x} [x^{2}] + \frac{d}{d x} [6]$

Step 1.1.2.4

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

$x^{2} (\frac{1}{3 x} (3 \cdot 1)) + \ln (3 x) \frac{d}{d x} [x^{2}] + \frac{d}{d x} [6]$

Step 1.1.2.5

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

$x^{2} (\frac{1}{3 x} (3 \cdot 1)) + \ln (3 x) (2 x) + \frac{d}{d x} [6]$

Step 1.1.2.6

Multiply $3$ by $1$ .

$x^{2} (\frac{1}{3 x} \cdot 3) + \ln (3 x) (2 x) + \frac{d}{d x} [6]$

Step 1.1.2.7

Combine $\frac{1}{3 x}$ and $3$ .

$x^{2} \frac{3}{3 x} + \ln (3 x) (2 x) + \frac{d}{d x} [6]$

Step 1.1.2.8

Cancel the common factor of $3$ .

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Step 1.1.2.8.1

Cancel the common factor.

$x^{2} \frac{3}{3 x} + \ln (3 x) (2 x) + \frac{d}{d x} [6]$

Step 1.1.2.8.2

Rewrite the expression.

$x^{2} \frac{1}{x} + \ln (3 x) (2 x) + \frac{d}{d x} [6]$

Step 1.1.2.9

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

$\frac{x^{2}}{x} + \ln (3 x) (2 x) + \frac{d}{d x} [6]$

Step 1.1.2.10

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

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Step 1.1.2.10.1

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

$\frac{x \cdot x}{x} + \ln (3 x) (2 x) + \frac{d}{d x} [6]$

Step 1.1.2.10.2

Cancel the common factors.

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Step 1.1.2.10.2.1

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

$\frac{x \cdot x}{x^{1}} + \ln (3 x) (2 x) + \frac{d}{d x} [6]$

Step 1.1.2.10.2.2

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

$\frac{x \cdot x}{x \cdot 1} + \ln (3 x) (2 x) + \frac{d}{d x} [6]$

Step 1.1.2.10.2.3

Cancel the common factor.

$\frac{x \cdot x}{x \cdot 1} + \ln (3 x) (2 x) + \frac{d}{d x} [6]$

Step 1.1.2.10.2.4

Rewrite the expression.

$\frac{x}{1} + \ln (3 x) (2 x) + \frac{d}{d x} [6]$

Step 1.1.2.10.2.5

Divide $x$ by $1$ .

$x + \ln (3 x) (2 x) + \frac{d}{d x} [6]$

Step 1.1.3

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$x + \ln (3 x) (2 x) + 0$

Step 1.1.4

Simplify.

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Step 1.1.4.1

Add $x + \ln (3 x) (2 x)$ and $0$ .

$x + \ln (3 x) (2 x)$

Step 1.1.4.2

Reorder terms.

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

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $2 x \ln (3 x) + x$ .

$2 x \ln (3 x) + x$

Step 2

Set the first derivative equal to $0$ then solve the equation $2 x \ln (3 x) + x = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$2 x \ln (3 x) + x = 0$

Step 2.2

Subtract $x$ from both sides of the equation.

$2 x \ln (3 x) = - x$

Step 2.3

Divide each term in $2 x \ln (3 x) = - x$ by $2 x$ and simplify.

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Step 2.3.1

Divide each term in $2 x \ln (3 x) = - x$ by $2 x$ .

$\frac{2 x \ln (3 x)}{2 x} = \frac{- x}{2 x}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $2$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{2 x \ln (3 x)}{2 x} = \frac{- x}{2 x}$

Step 2.3.2.1.2

Rewrite the expression.

$\frac{x \ln (3 x)}{x} = \frac{- x}{2 x}$

Step 2.3.2.2

Cancel the common factor of $x$ .

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Step 2.3.2.2.1

Cancel the common factor.

$\frac{x \ln (3 x)}{x} = \frac{- x}{2 x}$

Step 2.3.2.2.2

Divide $\ln (3 x)$ by $1$ .

$\ln (3 x) = \frac{- x}{2 x}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Cancel the common factor of $x$ .

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Step 2.3.3.1.1

Cancel the common factor.

$\ln (3 x) = \frac{- x}{2 x}$

Step 2.3.3.1.2

Rewrite the expression.

$\ln (3 x) = \frac{- 1}{2}$

Step 2.3.3.2

Move the negative in front of the fraction.

$\ln (3 x) = - \frac{1}{2}$

Step 2.4

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (3 x)} = e^{- \frac{1}{2}}$

Step 2.5

Rewrite $\ln (3 x) = - \frac{1}{2}$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{- \frac{1}{2}} = 3 x$

Step 2.6

Solve for $x$ .

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Step 2.6.1

Rewrite the equation as $3 x = e^{- \frac{1}{2}}$ .

$3 x = e^{- \frac{1}{2}}$

Step 2.6.2

Divide each term in $3 x = e^{- \frac{1}{2}}$ by $3$ and simplify.

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Step 2.6.2.1

Divide each term in $3 x = e^{- \frac{1}{2}}$ by $3$ .

$\frac{3 x}{3} = \frac{e^{- \frac{1}{2}}}{3}$

Step 2.6.2.2

Simplify the left side.

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Step 2.6.2.2.1

Cancel the common factor of $3$ .

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Step 2.6.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{e^{- \frac{1}{2}}}{3}$

Step 2.6.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{e^{- \frac{1}{2}}}{3}$

Step 2.6.2.3

Simplify the right side.

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Step 2.6.2.3.1

Move $e^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$x = \frac{1}{3 e^{\frac{1}{2}}}$

Step 3

Find the values where the derivative is undefined.

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Step 3.1

Set the argument in $\ln (3 x)$ less than or equal to $0$ to find where the expression is undefined.

$3 x \leq 0$

Step 3.2

Divide each term in $3 x \leq 0$ by $3$ and simplify.

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Step 3.2.1

Divide each term in $3 x \leq 0$ by $3$ .

$\frac{3 x}{3} \leq \frac{0}{3}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $3$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} \leq \frac{0}{3}$

Step 3.2.2.1.2

Divide $x$ by $1$ .

$x \leq \frac{0}{3}$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Divide $0$ by $3$ .

$x \leq 0$

Step 3.3

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x \leq 0$

$(- \infty, 0]$

$x \leq 0$

$(- \infty, 0]$

Step 4

Evaluate $x^{2} \ln (3 x) + 6$ at each $x$ value where the derivative is $0$ or undefined.

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Step 4.1

Evaluate at $x = \frac{1}{3 e^{\frac{1}{2}}}$ .

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Step 4.1.1

Substitute $\frac{1}{3 e^{\frac{1}{2}}}$ for $x$ .

${(\frac{1}{3 e^{\frac{1}{2}}})}^{2} \ln (3 (\frac{1}{3 e^{\frac{1}{2}}})) + 6$

Step 4.1.2

Simplify each term.

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Step 4.1.2.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 4.1.2.1.1

Apply the product rule to $\frac{1}{3 e^{\frac{1}{2}}}$ .

$\frac{1^{2}}{{(3 e^{\frac{1}{2}})}^{2}} \ln (3 (\frac{1}{3 e^{\frac{1}{2}}})) + 6$

Step 4.1.2.1.2

Apply the product rule to $3 e^{\frac{1}{2}}$ .

$\frac{1^{2}}{3^{2} {(e^{\frac{1}{2}})}^{2}} \ln (3 (\frac{1}{3 e^{\frac{1}{2}}})) + 6$

Step 4.1.2.2

One to any power is one.

$\frac{1}{3^{2} {(e^{\frac{1}{2}})}^{2}} \ln (3 (\frac{1}{3 e^{\frac{1}{2}}})) + 6$

Step 4.1.2.3

Simplify the denominator.

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Step 4.1.2.3.1

Raise $3$ to the power of $2$ .

$\frac{1}{9 {(e^{\frac{1}{2}})}^{2}} \ln (3 (\frac{1}{3 e^{\frac{1}{2}}})) + 6$

Step 4.1.2.3.2

Multiply the exponents in ${(e^{\frac{1}{2}})}^{2}$ .

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Step 4.1.2.3.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{1}{9 e^{\frac{1}{2} \cdot 2}} \ln (3 (\frac{1}{3 e^{\frac{1}{2}}})) + 6$

Step 4.1.2.3.2.2

Cancel the common factor of $2$ .

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Step 4.1.2.3.2.2.1

Cancel the common factor.

$\frac{1}{9 e^{\frac{1}{2} \cdot 2}} \ln (3 (\frac{1}{3 e^{\frac{1}{2}}})) + 6$

Step 4.1.2.3.2.2.2

Rewrite the expression.

$\frac{1}{9 e^{1}} \ln (3 (\frac{1}{3 e^{\frac{1}{2}}})) + 6$

Step 4.1.2.3.3

Simplify.

$\frac{1}{9 e} \ln (3 (\frac{1}{3 e^{\frac{1}{2}}})) + 6$

Step 4.1.2.4

Cancel the common factor of $3$ .

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Step 4.1.2.4.1

Cancel the common factor.

$\frac{1}{9 e} \ln (3 \frac{1}{3 e^{\frac{1}{2}}}) + 6$

Step 4.1.2.4.2

Rewrite the expression.

$\frac{1}{9 e} \ln (\frac{1}{e^{\frac{1}{2}}}) + 6$

Step 4.1.2.5

Move $e^{\frac{1}{2}}$ to the numerator using the negative exponent rule $\frac{1}{b^{n}} = b^{- n}$ .

$\frac{1}{9 e} \ln (e^{- \frac{1}{2}}) + 6$

Step 4.1.2.6

Expand $\ln (e^{- \frac{1}{2}})$ by moving $- \frac{1}{2}$ outside the logarithm.

$\frac{1}{9 e} (- \frac{1}{2} \ln (e)) + 6$

Step 4.1.2.7

The natural logarithm of $e$ is $1$ .

$\frac{1}{9 e} (- \frac{1}{2} \cdot 1) + 6$

Step 4.1.2.8

Multiply $- 1$ by $1$ .

$\frac{1}{9 e} (- \frac{1}{2}) + 6$

Step 4.1.2.9

Multiply $\frac{1}{9 e} (- \frac{1}{2})$ .

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Step 4.1.2.9.1

Multiply $\frac{1}{9 e}$ by $\frac{1}{2}$ .

$- \frac{1}{9 e \cdot 2} + 6$

Step 4.1.2.9.2

Multiply $2$ by $9$ .

$- \frac{1}{18 e} + 6$

Step 4.2

Evaluate at $x = 0$ .

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Step 4.2.1

Substitute $0$ for $x$ .

${(0)}^{2} \ln (3 (0)) + 6$

Step 4.2.2

Simplify.

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Step 4.2.2.1

Simplify each term.

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Step 4.2.2.1.1

Raising $0$ to any positive power yields $0$ .

$0 \ln (3 (0)) + 6$

Step 4.2.2.1.2

Rewrite $\ln (3 (0))$ as $\ln (3) + \ln (0)$ .

$0 (\ln (3) + \ln (0)) + 6$

Step 4.2.2.1.3

The natural logarithm of zero is undefined.

Undefined

$0 (\ln (3) + \ln (0)) + 6$

Step 4.2.2.2

The natural logarithm of zero is undefined.

Undefined

Step 4.3

List all of the points.

$(\frac{1}{3 e^{\frac{1}{2}}}, - \frac{1}{18 e} + 6)$

Step 5