Calculus Examples

$y = x^{3} - 4 x$ , $(2, 0)$

Step 1

Find the first derivative and evaluate at $x = 2$ and $y = 0$ to find the slope of the tangent line.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of $x^{3} - 4 x$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 4 x]$ .

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

Step 1.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$3 x^{2} + \frac{d}{d x} [- 4 x]$

Step 1.2

Evaluate $\frac{d}{d x} [- 4 x]$ .

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Step 1.2.1

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

$3 x^{2} - 4 \frac{d}{d x} [x]$

Step 1.2.2

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

$3 x^{2} - 4 \cdot 1$

Step 1.2.3

Multiply $- 4$ by $1$ .

$3 x^{2} - 4$

Step 1.3

Evaluate the derivative at $x = 2$ .

$3 {(2)}^{2} - 4$

Step 1.4

Simplify.

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

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

$3 \cdot 4 - 4$

Step 1.4.1.2

Multiply $3$ by $4$ .

$12 - 4$

Step 1.4.2

Subtract $4$ from $12$ .

$8$

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

$y - (0) = 8 \cdot (x - (2))$

Step 2.2

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

$y + 0 = 8 \cdot (x - 2)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Add $y$ and $0$ .

$y = 8 \cdot (x - 2)$

Step 2.3.2

Simplify $8 \cdot (x - 2)$ .

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Step 2.3.2.1

Apply the distributive property.

$y = 8 x + 8 \cdot - 2$

Step 2.3.2.2

Multiply $8$ by $- 2$ .

$y = 8 x - 16$

Step 3