Calculus Examples

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

Step 1

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

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Step 1.1

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

$3 x^{2}$

Step 1.2

Evaluate the derivative at $x = 1$ .

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

Step 1.3

Simplify.

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Step 1.3.1

One to any power is one.

$3 \cdot 1$

Step 1.3.2

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

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

Step 2.2

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

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

Step 2.3

Solve for $y$ .

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Step 2.3.1

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

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Step 2.3.1.1

Rewrite.

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

Step 2.3.1.2

Simplify by adding zeros.

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

Step 2.3.1.3

Apply the distributive property.

$y - 1 = 3 x + 3 \cdot - 1$

Step 2.3.1.4

Multiply $3$ by $- 1$ .

$y - 1 = 3 x - 3$

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 $1$ to both sides of the equation.

$y = 3 x - 3 + 1$

Step 2.3.2.2

Add $- 3$ and $1$ .

$y = 3 x - 2$

Step 3