Calculus Examples

$y = x^{3} - x$ at the point $(1, 0)$

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

Differentiate.

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Step 1.1.1

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

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

Step 1.1.2

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

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

Step 1.2

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

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Step 1.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

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

Step 1.2.2

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

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

Step 1.2.3

Multiply $- 1$ by $1$ .

$3 x^{2} - 1$

Step 1.3

Evaluate the derivative at $x = 1$ .

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

Step 1.4

Simplify.

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Step 1.4.1

Simplify each term.

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Step 1.4.1.1

One to any power is one.

$3 \cdot 1 - 1$

Step 1.4.1.2

Multiply $3$ by $1$ .

$3 - 1$

Step 1.4.2

Subtract $1$ from $3$ .

$2$

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

$y - (0) = 2 \cdot (x - (1))$

Step 2.2

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

$y + 0 = 2 \cdot (x - 1)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Add $y$ and $0$ .

$y = 2 \cdot (x - 1)$

Step 2.3.2

Simplify $2 \cdot (x - 1)$ .

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Step 2.3.2.1

Apply the distributive property.

$y = 2 x + 2 \cdot - 1$

Step 2.3.2.2

Multiply $2$ by $- 1$ .

$y = 2 x - 2$

Step 3