Calculus Examples

y = - \frac{1}{x^{3}}

$y = - \frac{1}{x^{3}}$ ,

(- 1, 1)

$(- 1, 1)$

Step 1

Find the first derivative and evaluate at

x = - 1

$x = - 1$ and

y = 1

$y = 1$ to find the slope of the tangent line.

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Step 1.1

Differentiate using the Product Rule which states that

\frac{d}{d x} [f (x) g (x)]

$\frac{d}{d x} [f (x) g (x)]$ is

f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]

$f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where

f (x) = - 1

$f (x) = - 1$ and

g (x) = \frac{1}{x^{3}}

$g (x) = \frac{1}{x^{3}}$ .

- \frac{d}{d x} [\frac{1}{x^{3}}] + \frac{1}{x^{3}} \frac{d}{d x} [- 1]

$- \frac{d}{d x} [\frac{1}{x^{3}}] + \frac{1}{x^{3}} \frac{d}{d x} [- 1]$

Step 1.2

Differentiate.

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Step 1.2.1

Rewrite

\frac{1}{x^{3}}

$\frac{1}{x^{3}}$ as

{(x^{3})}^{- 1}

${(x^{3})}^{- 1}$ .

- \frac{d}{d x} [{(x^{3})}^{- 1}] + \frac{1}{x^{3}} \frac{d}{d x} [- 1]

$- \frac{d}{d x} [{(x^{3})}^{- 1}] + \frac{1}{x^{3}} \frac{d}{d x} [- 1]$

Step 1.2.2

Multiply the exponents in

{(x^{3})}^{- 1}

${(x^{3})}^{- 1}$ .

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Step 1.2.2.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

- \frac{d}{d x} [x^{3 \cdot - 1}] + \frac{1}{x^{3}} \frac{d}{d x} [- 1]

$- \frac{d}{d x} [x^{3 \cdot - 1}] + \frac{1}{x^{3}} \frac{d}{d x} [- 1]$

Step 1.2.2.2

Multiply

3

$3$ by

- 1

$- 1$ .

- \frac{d}{d x} [x^{- 3}] + \frac{1}{x^{3}} \frac{d}{d x} [- 1]

$- \frac{d}{d x} [x^{- 3}] + \frac{1}{x^{3}} \frac{d}{d x} [- 1]$

- \frac{d}{d x} [x^{- 3}] + \frac{1}{x^{3}} \frac{d}{d x} [- 1]

$- \frac{d}{d x} [x^{- 3}] + \frac{1}{x^{3}} \frac{d}{d x} [- 1]$

Step 1.2.3

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = - 3

$n = - 3$ .

- (- 3 x^{- 4}) + \frac{1}{x^{3}} \frac{d}{d x} [- 1]

$- (- 3 x^{- 4}) + \frac{1}{x^{3}} \frac{d}{d x} [- 1]$

Step 1.2.4

Multiply

- 3

$- 3$ by

- 1

$- 1$ .

3 x^{- 4} + \frac{1}{x^{3}} \frac{d}{d x} [- 1]

$3 x^{- 4} + \frac{1}{x^{3}} \frac{d}{d x} [- 1]$

Step 1.2.5

Since

- 1

$- 1$ is constant with respect to

x

$x$ , the derivative of

- 1

$- 1$ with respect to

x

$x$ is

0

$0$ .

3 x^{- 4} + \frac{1}{x^{3}} \cdot 0

$3 x^{- 4} + \frac{1}{x^{3}} \cdot 0$

Step 1.2.6

Simplify the expression.

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Step 1.2.6.1

Multiply

\frac{1}{x^{3}}

$\frac{1}{x^{3}}$ by

0

$0$ .

3 x^{- 4} + 0

$3 x^{- 4} + 0$

Step 1.2.6.2

Add

3 x^{- 4}

$3 x^{- 4}$ and

0

$0$ .

3 x^{- 4}

$3 x^{- 4}$

3 x^{- 4}

$3 x^{- 4}$

3 x^{- 4}

$3 x^{- 4}$

Step 1.3

Simplify.

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Step 1.3.1

Rewrite the expression using the negative exponent rule

b^{- n} = \frac{1}{b^{n}}

$b^{- n} = \frac{1}{b^{n}}$ .

3 \frac{1}{x^{4}}

$3 \frac{1}{x^{4}}$

Step 1.3.2

Combine

3

$3$ and

\frac{1}{x^{4}}

$\frac{1}{x^{4}}$ .

\frac{3}{x^{4}}

$\frac{3}{x^{4}}$

\frac{3}{x^{4}}

$\frac{3}{x^{4}}$

Step 1.4

Evaluate the derivative at

x = - 1

$x = - 1$ .

\frac{3}{{(- 1)}^{4}}

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

Step 1.5

Simplify.

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Step 1.5.1

Raise

- 1

$- 1$ to the power of

4

$4$ .

\frac{3}{1}

$\frac{3}{1}$

Step 1.5.2

Divide

3

$3$ by

1

$1$ .

3

$3$

3

$3$

3

$3$

Step 2

Plug the slope and point values into the point-slope formula and solve for

y

$y$ .

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Step 2.1

Use the slope

3

$3$ and a given point

(- 1, 1)

$(- 1, 1)$ to substitute for

x_{1}

$x_{1}$ and

y_{1}

$y_{1}$ in the point-slope form

y - y_{1} = m (x - x_{1})

$y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

y - (1) = 3 \cdot (x - (- 1))

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

Step 2.2

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

y - 1 = 3 \cdot (x + 1)

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

Step 2.3

Solve for

y

$y$ .

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Step 2.3.1

Simplify

3 \cdot (x + 1)

$3 \cdot (x + 1)$ .

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Step 2.3.1.1

Rewrite.

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

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

Step 2.3.1.2

Simplify by adding zeros.

y - 1 = 3 \cdot (x + 1)

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

Step 2.3.1.3

Apply the distributive property.

y - 1 = 3 x + 3 \cdot 1

$y - 1 = 3 x + 3 \cdot 1$

Step 2.3.1.4

Multiply

3

$3$ by

1

$1$ .

y - 1 = 3 x + 3

$y - 1 = 3 x + 3$

y - 1 = 3 x + 3

$y - 1 = 3 x + 3$

Step 2.3.2

Move all terms not containing

y

$y$ to the right side of the equation.

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Step 2.3.2.1

Add

1

$1$ to both sides of the equation.

y = 3 x + 3 + 1

$y = 3 x + 3 + 1$

Step 2.3.2.2

Add

3

$3$ and

1

$1$ .

y = 3 x + 4

$y = 3 x + 4$

y = 3 x + 4

$y = 3 x + 4$

y = 3 x + 4

$y = 3 x + 4$

y = 3 x + 4

$y = 3 x + 4$

Step 3