Calculus Examples

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

Step 1

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

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Step 1.1

Substitute $1$ in for $x$ .

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

Step 1.2

Solve for $y$ .

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Step 1.2.1

Remove parentheses.

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

Step 1.2.2

Remove parentheses.

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

Step 1.2.3

Remove parentheses.

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

Step 1.2.4

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

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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.

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

Step 1.2.4.1.2

Multiply $- 1$ by $1$ .

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

Step 1.2.4.1.3

One to any power is one.

$y = \ln (4 - 1 + 2 \cdot 1)$

Step 1.2.4.1.4

Multiply $2$ by $1$ .

$y = \ln (4 - 1 + 2)$

Step 1.2.4.2

Simplify by adding and subtracting.

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Step 1.2.4.2.1

Subtract $1$ from $4$ .

$y = \ln (3 + 2)$

Step 1.2.4.2.2

Add $3$ and $2$ .

$y = \ln (5)$

Step 2

Find the first derivative and evaluate at $x = 1$ and $y = \ln (5)$ 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 $g (x) = 4 - x^{2} + 2 x^{4}$ .

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.1.3

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

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

Multiply $2$ by $- 1$ .

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

Simplify the expression.

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Step 2.2.9.1

Multiply $4$ by $2$ .

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

Step 2.2.9.2

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

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

Step 2.3

Evaluate the derivative at $x = 1$ .

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

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.

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

Step 2.4.1.2

Multiply $- 1$ by $1$ .

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

Step 2.4.1.3

One to any power is one.

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

Step 2.4.1.4

Multiply $2$ by $1$ .

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

Step 2.4.1.5

Subtract $1$ from $4$ .

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

Step 2.4.1.6

Add $3$ and $2$ .

$(- 2 \cdot 1 + 8 {(1)}^{3}) \frac{1}{5}$

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 $- 2$ by $1$ .

$(- 2 + 8 {(1)}^{3}) \frac{1}{5}$

Step 2.4.2.1.2

One to any power is one.

$(- 2 + 8 \cdot 1) \frac{1}{5}$

Step 2.4.2.1.3

Multiply $8$ by $1$ .

$(- 2 + 8) \frac{1}{5}$

Step 2.4.2.2

Combine fractions.

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Step 2.4.2.2.1

Add $- 2$ and $8$ .

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

Step 2.4.2.2.2

Combine $6$ and $\frac{1}{5}$ .

$\frac{6}{5}$

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 $\frac{6}{5}$ and a given point $(1, \ln (5))$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

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

Step 3.2

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

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

Step 3.3

Solve for $y$ .

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Step 3.3.1

Simplify $\frac{6}{5} \cdot (x - 1)$ .

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Step 3.3.1.1

Rewrite.

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

Step 3.3.1.2

Simplify by adding zeros.

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

Step 3.3.1.3

Apply the distributive property.

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

Step 3.3.1.4

Combine $\frac{6}{5}$ and $x$ .

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

Step 3.3.1.5

Multiply $\frac{6}{5} \cdot - 1$ .

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Step 3.3.1.5.1

Combine $\frac{6}{5}$ and $- 1$ .

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

Step 3.3.1.5.2

Multiply $6$ by $- 1$ .

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

Step 3.3.1.6

Move the negative in front of the fraction.

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

Step 3.3.2

Add $\ln (5)$ to both sides of the equation.

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

Step 3.3.3

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

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Step 3.3.3.1

To write $\ln (5)$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

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

Step 3.3.3.2

Combine $\ln (5)$ and $\frac{5}{5}$ .

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

Step 3.3.3.3

Combine the numerators over the common denominator.

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

Step 3.3.3.4

Simplify the numerator.

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Step 3.3.3.4.1

Multiply $\ln (5) \cdot 5$ .

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Step 3.3.3.4.1.1

Reorder $\ln (5)$ and $5$ .

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

Step 3.3.3.4.1.2

Simplify $5 \cdot \ln (5)$ by moving $5$ inside the logarithm.

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

Step 3.3.3.4.2

Raise $5$ to the power of $5$ .

$y = \frac{6 x}{5} + \frac{- 6 + \ln (3125)}{5}$

Step 3.3.3.5

Rewrite $- 6$ as $- 1 (6)$ .

$y = \frac{6 x}{5} + \frac{- 1 (6) + \ln (3125)}{5}$

Step 3.3.3.6

Factor $- 1$ out of $\ln (3125)$ .

$y = \frac{6 x}{5} + \frac{- 1 (6) - 1 (- \ln (3125))}{5}$

Step 3.3.3.7

Factor $- 1$ out of $- 1 (6) - 1 (- \ln (3125))$ .

$y = \frac{6 x}{5} + \frac{- 1 (6 - \ln (3125))}{5}$

Step 3.3.3.8

Move the negative in front of the fraction.

$y = \frac{6 x}{5} - \frac{6 - \ln (3125)}{5}$

Step 3.3.3.9

Reorder terms.

$y = \frac{6}{5} x - \frac{6 - \ln (3125)}{5}$

Step 4