Calculus Examples

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

Step 1

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

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Step 1.1

Substitute $1$ in for $x$ .

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

Step 1.2

Solve for $y$ .

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Step 1.2.1

Remove parentheses.

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

Step 1.2.2

Remove parentheses.

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

Step 1.2.3

Remove parentheses.

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

Step 1.2.4

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

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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.

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

Step 1.2.4.1.2

Multiply $- 1$ by $1$ .

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

Step 1.2.4.1.3

One to any power is one.

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

Step 1.2.4.1.4

Multiply $2$ by $1$ .

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

Step 1.2.4.2

Simplify by adding and subtracting.

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Step 1.2.4.2.1

Subtract $1$ from $2$ .

$y = \ln (1 + 2)$

Step 1.2.4.2.2

Add $1$ and $2$ .

$y = \ln (3)$

Step 2

Find the first derivative and evaluate at $x = 1$ and $y = \ln (3)$ 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 $g (x) = 2 - x^{2} + 2 x^{4}$ .

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.1.3

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

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

Multiply $2$ by $- 1$ .

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

Step 2.2.7

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

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

Step 2.2.8

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

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

Step 2.2.9

Simplify the expression.

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Step 2.2.9.1

Multiply $4$ by $2$ .

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

Step 2.2.9.2

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

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

Step 2.3

Evaluate the derivative at $x = 1$ .

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

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.

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

Step 2.4.1.2

Multiply $- 1$ by $1$ .

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

Step 2.4.1.3

One to any power is one.

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

Step 2.4.1.4

Multiply $2$ by $1$ .

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

Step 2.4.1.5

Subtract $1$ from $2$ .

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

Step 2.4.1.6

Add $1$ and $2$ .

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

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 $1$ .

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

Step 2.4.2.1.2

One to any power is one.

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

Step 2.4.2.1.3

Multiply $8$ by $1$ .

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

Step 2.4.2.2

Reduce the expression by cancelling the common factors.

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Step 2.4.2.2.1

Add $- 2$ and $8$ .

$6 (\frac{1}{3})$

Step 2.4.2.2.2

Cancel the common factor of $3$ .

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Step 2.4.2.2.2.1

Factor $3$ out of $6$ .

$3 (2) \frac{1}{3}$

Step 2.4.2.2.2.2

Cancel the common factor.

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

Step 2.4.2.2.2.3

Rewrite the expression.

$2$

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 $2$ and a given point $(1, \ln (3))$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (\ln (3)) = 2 \cdot (x - (1))$

Step 3.2

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

$y - \ln (3) = 2 \cdot (x - 1)$

Step 3.3

Solve for $y$ .

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Step 3.3.1

Simplify $2 \cdot (x - 1)$ .

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Step 3.3.1.1

Rewrite.

$y - \ln (3) = 0 + 0 + 2 \cdot (x - 1)$

Step 3.3.1.2

Simplify by adding zeros.

$y - \ln (3) = 2 \cdot (x - 1)$

Step 3.3.1.3

Apply the distributive property.

$y - \ln (3) = 2 x + 2 \cdot - 1$

Step 3.3.1.4

Multiply $2$ by $- 1$ .

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

Step 3.3.2

Add $\ln (3)$ to both sides of the equation.

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

Step 4