Calculus Examples

$y = \frac{1 + x}{8 + e^{x}}$ , $(0, \frac{1}{9})$

Step 1

Find the first derivative and evaluate at $x = 0$ and $y = \frac{1}{9}$ to find the slope of the tangent line.

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Step 1.1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = 1 + x$ and $g (x) = 8 + e^{x}$ .

$\frac{(8 + e^{x}) \frac{d}{d x} [1 + x] - (1 + x) \frac{d}{d x} [8 + e^{x}]}{{(8 + e^{x})}^{2}}$

Step 1.2

Differentiate.

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Step 1.2.1

By the Sum Rule, the derivative of $1 + x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [x]$ .

$\frac{(8 + e^{x}) (\frac{d}{d x} [1] + \frac{d}{d x} [x]) - (1 + x) \frac{d}{d x} [8 + e^{x}]}{{(8 + e^{x})}^{2}}$

Step 1.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\frac{(8 + e^{x}) (0 + \frac{d}{d x} [x]) - (1 + x) \frac{d}{d x} [8 + e^{x}]}{{(8 + e^{x})}^{2}}$

Step 1.2.3

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

$\frac{(8 + e^{x}) \frac{d}{d x} [x] - (1 + x) \frac{d}{d x} [8 + e^{x}]}{{(8 + e^{x})}^{2}}$

Step 1.2.4

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

$\frac{(8 + e^{x}) \cdot 1 - (1 + x) \frac{d}{d x} [8 + e^{x}]}{{(8 + e^{x})}^{2}}$

Step 1.2.5

Multiply $8 + e^{x}$ by $1$ .

$\frac{8 + e^{x} - (1 + x) \frac{d}{d x} [8 + e^{x}]}{{(8 + e^{x})}^{2}}$

Step 1.2.6

By the Sum Rule, the derivative of $8 + e^{x}$ with respect to $x$ is $\frac{d}{d x} [8] + \frac{d}{d x} [e^{x}]$ .

$\frac{8 + e^{x} - (1 + x) (\frac{d}{d x} [8] + \frac{d}{d x} [e^{x}])}{{(8 + e^{x})}^{2}}$

Step 1.2.7

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$\frac{8 + e^{x} - (1 + x) (0 + \frac{d}{d x} [e^{x}])}{{(8 + e^{x})}^{2}}$

Step 1.2.8

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

$\frac{8 + e^{x} - (1 + x) \frac{d}{d x} [e^{x}]}{{(8 + e^{x})}^{2}}$

Step 1.3

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

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

Step 1.4

Simplify.

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Step 1.4.1

Apply the distributive property.

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

Step 1.4.2

Apply the distributive property.

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

Step 1.4.3

Simplify the numerator.

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Step 1.4.3.1

Simplify each term.

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Step 1.4.3.1.1

Multiply $- 1$ by $1$ .

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

Step 1.4.3.1.2

Rewrite $- 1 e^{x}$ as $- e^{x}$ .

$\frac{8 + e^{x} - e^{x} - x e^{x}}{{(8 + e^{x})}^{2}}$

Step 1.4.3.2

Combine the opposite terms in $8 + e^{x} - e^{x} - x e^{x}$ .

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Step 1.4.3.2.1

Subtract $e^{x}$ from $e^{x}$ .

$\frac{8 + 0 - x e^{x}}{{(8 + e^{x})}^{2}}$

Step 1.4.3.2.2

Add $8$ and $0$ .

$\frac{8 - x e^{x}}{{(8 + e^{x})}^{2}}$

Step 1.4.4

Reorder terms.

$\frac{- e^{x} x + 8}{{(8 + e^{x})}^{2}}$

Step 1.4.5

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

$\frac{- (e^{x} x) + 8}{{(8 + e^{x})}^{2}}$

Step 1.4.6

Rewrite $8$ as $- 1 (- 8)$ .

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

Step 1.4.7

Factor $- 1$ out of $- (e^{x} x) - 1 (- 8)$ .

$\frac{- (e^{x} x - 8)}{{(8 + e^{x})}^{2}}$

Step 1.4.8

Rewrite $- (e^{x} x - 8)$ as $- 1 (e^{x} x - 8)$ .

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

Step 1.4.9

Move the negative in front of the fraction.

$- \frac{e^{x} x - 8}{{(8 + e^{x})}^{2}}$

Step 1.4.10

Reorder factors in $- \frac{e^{x} x - 8}{{(8 + e^{x})}^{2}}$ .

$- \frac{x e^{x} - 8}{{(8 + e^{x})}^{2}}$

Step 1.5

Evaluate the derivative at $x = 0$ .

$- \frac{(0) \cdot e^{0} - 8}{{(8 + e^{0})}^{2}}$

Step 1.6

Simplify.

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Step 1.6.1

Simplify the numerator.

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Step 1.6.1.1

Rewrite $0 \cdot e^{0}$ as ${(0 \cdot e^{0})}^{3}$ .

$- \frac{{(0 \cdot e^{0})}^{3} - 8}{{(8 + e^{0})}^{2}}$

Step 1.6.1.2

Rewrite $8$ as $2^{3}$ .

$- \frac{{(0 \cdot e^{0})}^{3} - 2^{3}}{{(8 + e^{0})}^{2}}$

Step 1.6.1.3

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = 0 \cdot e^{0}$ and $b = 2$ .

$- \frac{(0 \cdot e^{0} - 2) ({(0 \cdot e^{0})}^{2} + 0 \cdot e^{0} \cdot 2 + 2^{2})}{{(8 + e^{0})}^{2}}$

Step 1.6.1.4

Simplify.

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Step 1.6.1.4.1

Anything raised to $0$ is $1$ .

$- \frac{(0 \cdot 1 - 2) ({(0 \cdot e^{0})}^{2} + 0 \cdot e^{0} \cdot 2 + 2^{2})}{{(8 + e^{0})}^{2}}$

Step 1.6.1.4.2

Multiply $0$ by $1$ .

$- \frac{(0 - 2) ({(0 \cdot e^{0})}^{2} + 0 \cdot e^{0} \cdot 2 + 2^{2})}{{(8 + e^{0})}^{2}}$

Step 1.6.1.4.3

Subtract $2$ from $0$ .

$- \frac{- 2 ({(0 \cdot e^{0})}^{2} + 0 \cdot e^{0} \cdot 2 + 2^{2})}{{(8 + e^{0})}^{2}}$

Step 1.6.1.4.4

Factor using the perfect square rule.

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Step 1.6.1.4.4.1

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$0 \cdot e^{0} \cdot 2 = 2 \cdot (0 \cdot e^{0}) \cdot 2$

Step 1.6.1.4.4.2

Rewrite the polynomial.

$- \frac{- 2 ({(0 \cdot e^{0})}^{2} + 2 \cdot (0 \cdot e^{0}) \cdot 2 + 2^{2})}{{(8 + e^{0})}^{2}}$

Step 1.6.1.4.4.3

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 0 \cdot e^{0}$ and $b = 2$ .

$- \frac{- 2 {(0 \cdot e^{0} + 2)}^{2}}{{(8 + e^{0})}^{2}}$

Step 1.6.1.4.5

Anything raised to $0$ is $1$ .

$- \frac{- 2 {(0 \cdot 1 + 2)}^{2}}{{(8 + e^{0})}^{2}}$

Step 1.6.1.4.6

Multiply $0$ by $1$ .

$- \frac{- 2 {(0 + 2)}^{2}}{{(8 + e^{0})}^{2}}$

Step 1.6.1.4.7

Add $0$ and $2$ .

$- \frac{- 2 \cdot 2^{2}}{{(8 + e^{0})}^{2}}$

Step 1.6.1.5

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

$- \frac{- 2 \cdot 4}{{(8 + e^{0})}^{2}}$

Step 1.6.2

Simplify the denominator.

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Step 1.6.2.1

Rewrite $8$ as $2^{3}$ .

$- \frac{- 2 \cdot 4}{{(2^{3} + e^{0})}^{2}}$

Step 1.6.2.2

Rewrite $e^{0}$ as ${(e^{0})}^{3}$ .

$- \frac{- 2 \cdot 4}{{(2^{3} + {(e^{0})}^{3})}^{2}}$

Step 1.6.2.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 2$ and $b = e^{0}$ .

$- \frac{- 2 \cdot 4}{{((2 + e^{0}) (2^{2} - 2 e^{0} + {(e^{0})}^{2}))}^{2}}$

Step 1.6.2.4

Simplify.

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Step 1.6.2.4.1

Anything raised to $0$ is $1$ .

$- \frac{- 2 \cdot 4}{{((2 + 1) (2^{2} - 2 e^{0} + {(e^{0})}^{2}))}^{2}}$

Step 1.6.2.4.2

Add $2$ and $1$ .

$- \frac{- 2 \cdot 4}{{(3 (2^{2} - 2 e^{0} + {(e^{0})}^{2}))}^{2}}$

Step 1.6.2.4.3

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

$- \frac{- 2 \cdot 4}{{(3 (4 - 2 e^{0} + {(e^{0})}^{2}))}^{2}}$

Step 1.6.2.4.4

Anything raised to $0$ is $1$ .

$- \frac{- 2 \cdot 4}{{(3 (4 - 2 \cdot 1 + {(e^{0})}^{2}))}^{2}}$

Step 1.6.2.4.5

Multiply $- 2$ by $1$ .

$- \frac{- 2 \cdot 4}{{(3 (4 - 2 + {(e^{0})}^{2}))}^{2}}$

Step 1.6.2.4.6

Multiply the exponents in ${(e^{0})}^{2}$ .

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Step 1.6.2.4.6.1

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

$- \frac{- 2 \cdot 4}{{(3 (4 - 2 + e^{0 \cdot 2}))}^{2}}$

Step 1.6.2.4.6.2

Multiply $0$ by $2$ .

$- \frac{- 2 \cdot 4}{{(3 (4 - 2 + e^{0}))}^{2}}$

Step 1.6.2.4.7

Anything raised to $0$ is $1$ .

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

Step 1.6.2.4.8

Subtract $2$ from $4$ .

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

Step 1.6.2.4.9

Add $2$ and $1$ .

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

Step 1.6.2.5

Apply the product rule to $3 \cdot 3$ .

$- \frac{- 2 \cdot 4}{3^{2} \cdot 3^{2}}$

Step 1.6.2.6

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

$- \frac{- 2 \cdot 4}{9 \cdot 3^{2}}$

Step 1.6.2.7

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

$- \frac{- 2 \cdot 4}{9 \cdot 9}$

Step 1.6.3

Simplify the expression.

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Step 1.6.3.1

Multiply $- 2$ by $4$ .

$- \frac{- 8}{9 \cdot 9}$

Step 1.6.3.2

Multiply $9$ by $9$ .

$- \frac{- 8}{81}$

Step 1.6.3.3

Move the negative in front of the fraction.

$- - \frac{8}{81}$

Step 1.6.4

Multiply $- - \frac{8}{81}$ .

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Step 1.6.4.1

Multiply $- 1$ by $- 1$ .

$1 (\frac{8}{81})$

Step 1.6.4.2

Multiply $\frac{8}{81}$ by $1$ .

$\frac{8}{81}$

Step 2

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

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Step 2.1

Use the slope $\frac{8}{81}$ and a given point $(0, \frac{1}{9})$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (\frac{1}{9}) = \frac{8}{81} \cdot (x - (0))$

Step 2.2

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

$y - \frac{1}{9} = \frac{8}{81} \cdot (x + 0)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Simplify $\frac{8}{81} \cdot (x + 0)$ .

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Step 2.3.1.1

Add $x$ and $0$ .

$y - \frac{1}{9} = \frac{8}{81} \cdot x$

Step 2.3.1.2

Combine $\frac{8}{81}$ and $x$ .

$y - \frac{1}{9} = \frac{8 x}{81}$

Step 2.3.2

Add $\frac{1}{9}$ to both sides of the equation.

$y = \frac{8 x}{81} + \frac{1}{9}$

Step 2.3.3

Reorder terms.

$y = \frac{8}{81} x + \frac{1}{9}$

Step 3