Calculus Examples

$f (x) = \ln (13 - 9 x^{2} + 2 x^{4})$ , $x = 1$

Step 1

Find the corresponding $y$ -value to $x = 1$ .

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Step 1.1

Substitute $1$ in for $x$ .

$y = \ln (13 - 9 {(1)}^{2} + 2 {(1)}^{4})$

Step 1.2

Solve for $y$ .

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Step 1.2.1

Remove parentheses.

$y = \ln (13 - 9 \cdot 1^{2} + 2 {(1)}^{4})$

Step 1.2.2

Remove parentheses.

$y = \ln (13 - 9 \cdot 1^{2} + 2 \cdot 1^{4})$

Step 1.2.3

Remove parentheses.

$y = \ln (13 - 9 {(1)}^{2} + 2 {(1)}^{4})$

Step 1.2.4

Simplify $\ln (13 - 9 {(1)}^{2} + 2 {(1)}^{4})$ .

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Step 1.2.4.1

Simplify each term.

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Step 1.2.4.1.1

One to any power is one.

$y = \ln (13 - 9 \cdot 1 + 2 {(1)}^{4})$

Step 1.2.4.1.2

Multiply $- 9$ by $1$ .

$y = \ln (13 - 9 + 2 {(1)}^{4})$

Step 1.2.4.1.3

One to any power is one.

$y = \ln (13 - 9 + 2 \cdot 1)$

Step 1.2.4.1.4

Multiply $2$ by $1$ .

$y = \ln (13 - 9 + 2)$

Step 1.2.4.2

Simplify by adding and subtracting.

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Step 1.2.4.2.1

Subtract $9$ from $13$ .

$y = \ln (4 + 2)$

Step 1.2.4.2.2

Add $4$ and $2$ .

$y = \ln (6)$

Step 2

Find the first derivative and evaluate at $x = 1$ and $y = \ln (6)$ to find the slope of the tangent line.

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Step 2.1

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 $g (x) = 13 - 9 x^{2} + 2 x^{4}$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $13 - 9 x^{2} + 2 x^{4}$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [13 - 9 x^{2} + 2 x^{4}]$

Step 2.1.2

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

$\frac{1}{u} \frac{d}{d x} [13 - 9 x^{2} + 2 x^{4}]$

Step 2.1.3

Replace all occurrences of $u$ with $13 - 9 x^{2} + 2 x^{4}$ .

$\frac{1}{13 - 9 x^{2} + 2 x^{4}} \frac{d}{d x} [13 - 9 x^{2} + 2 x^{4}]$

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $13 - 9 x^{2} + 2 x^{4}$ with respect to $x$ is $\frac{d}{d x} [13] + \frac{d}{d x} [- 9 x^{2}] + \frac{d}{d x} [2 x^{4}]$ .

$\frac{1}{13 - 9 x^{2} + 2 x^{4}} (\frac{d}{d x} [13] + \frac{d}{d x} [- 9 x^{2}] + \frac{d}{d x} [2 x^{4}])$

Step 2.2.2

Since $13$ is constant with respect to $x$ , the derivative of $13$ with respect to $x$ is $0$ .

$\frac{1}{13 - 9 x^{2} + 2 x^{4}} (0 + \frac{d}{d x} [- 9 x^{2}] + \frac{d}{d x} [2 x^{4}])$

Step 2.2.3

Add $0$ and $\frac{d}{d x} [- 9 x^{2}]$ .

$\frac{1}{13 - 9 x^{2} + 2 x^{4}} (\frac{d}{d x} [- 9 x^{2}] + \frac{d}{d x} [2 x^{4}])$

Step 2.2.4

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x^{2}$ with respect to $x$ is $- 9 \frac{d}{d x} [x^{2}]$ .

$\frac{1}{13 - 9 x^{2} + 2 x^{4}} (- 9 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [2 x^{4}])$

Step 2.2.5

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

$\frac{1}{13 - 9 x^{2} + 2 x^{4}} (- 9 (2 x) + \frac{d}{d x} [2 x^{4}])$

Step 2.2.6

Multiply $2$ by $- 9$ .

$\frac{1}{13 - 9 x^{2} + 2 x^{4}} (- 18 x + \frac{d}{d x} [2 x^{4}])$

Step 2.2.7

Since $2$ is constant with respect to $x$ , the derivative of $2 x^{4}$ with respect to $x$ is $2 \frac{d}{d x} [x^{4}]$ .

$\frac{1}{13 - 9 x^{2} + 2 x^{4}} (- 18 x + 2 \frac{d}{d x} [x^{4}])$

Step 2.2.8

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$\frac{1}{13 - 9 x^{2} + 2 x^{4}} (- 18 x + 2 (4 x^{3}))$

Step 2.2.9

Simplify the expression.

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Step 2.2.9.1

Multiply $4$ by $2$ .

$\frac{1}{13 - 9 x^{2} + 2 x^{4}} (- 18 x + 8 x^{3})$

Step 2.2.9.2

Reorder the factors of $\frac{1}{13 - 9 x^{2} + 2 x^{4}} (- 18 x + 8 x^{3})$ .

$(- 18 x + 8 x^{3}) \frac{1}{13 - 9 x^{2} + 2 x^{4}}$

Step 2.3

Evaluate the derivative at $x = 1$ .

$(- 18 \cdot 1 + 8 {(1)}^{3}) (\frac{1}{13 - 9 {(1)}^{2} + 2 {(1)}^{4}})$

Step 2.4

Simplify.

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Step 2.4.1

Simplify the denominator.

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Step 2.4.1.1

One to any power is one.

$(- 18 \cdot 1 + 8 {(1)}^{3}) \frac{1}{13 - 9 \cdot 1 + 2 {(1)}^{4}}$

Step 2.4.1.2

Multiply $- 9$ by $1$ .

$(- 18 \cdot 1 + 8 {(1)}^{3}) \frac{1}{13 - 9 + 2 {(1)}^{4}}$

Step 2.4.1.3

One to any power is one.

$(- 18 \cdot 1 + 8 {(1)}^{3}) \frac{1}{13 - 9 + 2 \cdot 1}$

Step 2.4.1.4

Multiply $2$ by $1$ .

$(- 18 \cdot 1 + 8 {(1)}^{3}) \frac{1}{13 - 9 + 2}$

Step 2.4.1.5

Subtract $9$ from $13$ .

$(- 18 \cdot 1 + 8 {(1)}^{3}) \frac{1}{4 + 2}$

Step 2.4.1.6

Add $4$ and $2$ .

$(- 18 \cdot 1 + 8 {(1)}^{3}) \frac{1}{6}$

Step 2.4.2

Simplify terms.

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Step 2.4.2.1

Simplify each term.

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Step 2.4.2.1.1

Multiply $- 18$ by $1$ .

$(- 18 + 8 {(1)}^{3}) \frac{1}{6}$

Step 2.4.2.1.2

One to any power is one.

$(- 18 + 8 \cdot 1) \frac{1}{6}$

Step 2.4.2.1.3

Multiply $8$ by $1$ .

$(- 18 + 8) \frac{1}{6}$

Step 2.4.2.2

Simplify terms.

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Step 2.4.2.2.1

Add $- 18$ and $8$ .

$- 10 (\frac{1}{6})$

Step 2.4.2.2.2

Cancel the common factor of $2$ .

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Step 2.4.2.2.2.1

Factor $2$ out of $- 10$ .

$2 (- 5) \frac{1}{6}$

Step 2.4.2.2.2.2

Factor $2$ out of $6$ .

$2 \cdot - 5 \frac{1}{2 \cdot 3}$

Step 2.4.2.2.2.3

Cancel the common factor.

$2 \cdot - 5 \frac{1}{2 \cdot 3}$

Step 2.4.2.2.2.4

Rewrite the expression.

$- 5 (\frac{1}{3})$

Step 2.4.2.2.3

Combine $- 5$ and $\frac{1}{3}$ .

$\frac{- 5}{3}$

Step 2.4.2.2.4

Move the negative in front of the fraction.

$- \frac{5}{3}$

Step 3

Plug the slope and point values into the point-slope formula and solve for $y$ .

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Step 3.1

Use the slope $- \frac{5}{3}$ and a given point $(1, \ln (6))$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (\ln (6)) = - \frac{5}{3} \cdot (x - (1))$

Step 3.2

Simplify the equation and keep it in point-slope form.

$y - \ln (6) = - \frac{5}{3} \cdot (x - 1)$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Simplify $- \frac{5}{3} \cdot (x - 1)$ .

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Step 3.3.1.1

Rewrite.

$y - \ln (6) = 0 + 0 - \frac{5}{3} \cdot (x - 1)$

Step 3.3.1.2

Simplify by adding zeros.

$y - \ln (6) = - \frac{5}{3} \cdot (x - 1)$

Step 3.3.1.3

Apply the distributive property.

$y - \ln (6) = - \frac{5}{3} x - \frac{5}{3} \cdot - 1$

Step 3.3.1.4

Combine $x$ and $\frac{5}{3}$ .

$y - \ln (6) = - \frac{x \cdot 5}{3} - \frac{5}{3} \cdot - 1$

Step 3.3.1.5

Multiply $- \frac{5}{3} \cdot - 1$ .

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Step 3.3.1.5.1

Multiply $- 1$ by $- 1$ .

$y - \ln (6) = - \frac{x \cdot 5}{3} + 1 (\frac{5}{3})$

Step 3.3.1.5.2

Multiply $\frac{5}{3}$ by $1$ .

$y - \ln (6) = - \frac{x \cdot 5}{3} + \frac{5}{3}$

Step 3.3.1.6

Move $5$ to the left of $x$ .

$y - \ln (6) = - \frac{5 x}{3} + \frac{5}{3}$

Step 3.3.2

Add $\ln (6)$ to both sides of the equation.

$y = - \frac{5 x}{3} + \frac{5}{3} + \ln (6)$

Step 3.3.3

Write in $y = m x + b$ form.

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Step 3.3.3.1

To write $\ln (6)$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = - \frac{5 x}{3} + \frac{5}{3} + \ln (6) \cdot \frac{3}{3}$

Step 3.3.3.2

Combine $\ln (6)$ and $\frac{3}{3}$ .

$y = - \frac{5 x}{3} + \frac{5}{3} + \frac{\ln (6) \cdot 3}{3}$

Step 3.3.3.3

Combine the numerators over the common denominator.

$y = - \frac{5 x}{3} + \frac{5 + \ln (6) \cdot 3}{3}$

Step 3.3.3.4

Simplify the numerator.

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Step 3.3.3.4.1

Multiply $\ln (6) \cdot 3$ .

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Step 3.3.3.4.1.1

Reorder $\ln (6)$ and $3$ .

$y = - \frac{5 x}{3} + \frac{5 + 3 \cdot \ln (6)}{3}$

Step 3.3.3.4.1.2

Simplify $3 \cdot \ln (6)$ by moving $3$ inside the logarithm.

$y = - \frac{5 x}{3} + \frac{5 + \ln (6^{3})}{3}$

Step 3.3.3.4.2

Raise $6$ to the power of $3$ .

$y = - \frac{5 x}{3} + \frac{5 + \ln (216)}{3}$

Step 3.3.3.5

Reorder terms.

$y = - (\frac{5}{3} x) + \frac{5 + \ln (216)}{3}$

Step 3.3.3.6

Remove parentheses.

$y = - \frac{5}{3} x + \frac{5 + \ln (216)}{3}$

Step 4