Calculus Examples

$f (x) = 2 x^{3}$ at $(- 2, - 16)$

Step 1

Find the first derivative and evaluate at $x = - 2$ and $y = - 16$ to find the slope of the tangent line.

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Step 1.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x^{3}$ with respect to $x$ is $2 \frac{d}{d x} [x^{3}]$ .

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

Step 1.2

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

$2 (3 x^{2})$

Step 1.3

Multiply $3$ by $2$ .

$6 x^{2}$

Step 1.4

Evaluate the derivative at $x = - 2$ .

$6 {(- 2)}^{2}$

Step 1.5

Simplify.

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Step 1.5.1

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

$6 \cdot 4$

Step 1.5.2

Multiply $6$ by $4$ .

$24$

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 $24$ and a given point $(- 2, - 16)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (- 16) = 24 \cdot (x - (- 2))$

Step 2.2

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

$y + 16 = 24 \cdot (x + 2)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Simplify $24 \cdot (x + 2)$ .

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Step 2.3.1.1

Rewrite.

$y + 16 = 0 + 0 + 24 \cdot (x + 2)$

Step 2.3.1.2

Simplify by adding zeros.

$y + 16 = 24 \cdot (x + 2)$

Step 2.3.1.3

Apply the distributive property.

$y + 16 = 24 x + 24 \cdot 2$

Step 2.3.1.4

Multiply $24$ by $2$ .

$y + 16 = 24 x + 48$

Step 2.3.2

Move all terms not containing $y$ to the right side of the equation.

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Step 2.3.2.1

Subtract $16$ from both sides of the equation.

$y = 24 x + 48 - 16$

Step 2.3.2.2

Subtract $16$ from $48$ .

$y = 24 x + 32$

Step 3