Calculus Examples

$f (x) = x^{3} + 1$ , $(- 1, 0)$

Step 1

Find the first derivative and evaluate at $x = - 1$ and $y = 0$ to find the slope of the tangent line.

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Step 1.1

By the Sum Rule, the derivative of $x^{3} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [1]$ .

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

Step 1.2

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

$3 x^{2} + \frac{d}{d x} [1]$

Step 1.3

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

$3 x^{2} + 0$

Step 1.4

Add $3 x^{2}$ and $0$ .

$3 x^{2}$

Step 1.5

Evaluate the derivative at $x = - 1$ .

$3 {(- 1)}^{2}$

Step 1.6

Simplify.

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Step 1.6.1

Raise $- 1$ to the power of $2$ .

$3 \cdot 1$

Step 1.6.2

Multiply $3$ by $1$ .

$3$

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 $3$ and a given point $(- 1, 0)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

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

Step 2.2

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

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

Step 2.3

Solve for $y$ .

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Step 2.3.1

Add $y$ and $0$ .

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

Step 2.3.2

Simplify $3 \cdot (x + 1)$ .

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Step 2.3.2.1

Apply the distributive property.

$y = 3 x + 3 \cdot 1$

Step 2.3.2.2

Multiply $3$ by $1$ .

$y = 3 x + 3$

Step 3