Calculus Examples

$f (x) = x^{2} + 1$ , $(2, 5)$

Step 1

Find the first derivative and evaluate at $x = 2$ and $y = 5$ to find the slope of the tangent line.

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Step 1.1

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

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

Step 1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 1.3

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

$2 x + 0$

Step 1.4

Add $2 x$ and $0$ .

$2 x$

Step 1.5

Evaluate the derivative at $x = 2$ .

$2 (2)$

Step 1.6

Multiply $2$ by $2$ .

$4$

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 $4$ and a given point $(2, 5)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (5) = 4 \cdot (x - (2))$

Step 2.2

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

$y - 5 = 4 \cdot (x - 2)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Simplify $4 \cdot (x - 2)$ .

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Step 2.3.1.1

Rewrite.

$y - 5 = 0 + 0 + 4 \cdot (x - 2)$

Step 2.3.1.2

Simplify by adding zeros.

$y - 5 = 4 \cdot (x - 2)$

Step 2.3.1.3

Apply the distributive property.

$y - 5 = 4 x + 4 \cdot - 2$

Step 2.3.1.4

Multiply $4$ by $- 2$ .

$y - 5 = 4 x - 8$

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 $5$ to both sides of the equation.

$y = 4 x - 8 + 5$

Step 2.3.2.2

Add $- 8$ and $5$ .

$y = 4 x - 3$

Step 3