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

Add $16$ to both sides of the equation.

$y = - 24 x - 48 + 16$

Step 2.3.2.2

Add $- 48$ and $16$ .

$y = - 24 x - 32$

Step 3