Calculus Examples

$f (x) = {(5 x + 1)}^{2}$ ;, $(0, 1)$

Step 1

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

Tap for more steps...

Step 1.1

Rewrite ${(5 x + 1)}^{2}$ as $(5 x + 1) (5 x + 1)$ .

$\frac{d}{d x} [(5 x + 1) (5 x + 1)]$

Step 1.2

Expand $(5 x + 1) (5 x + 1)$ using the FOIL Method.

Tap for more steps...

Step 1.2.1

Apply the distributive property.

$\frac{d}{d x} [5 x (5 x + 1) + 1 (5 x + 1)]$

Step 1.2.2

Apply the distributive property.

$\frac{d}{d x} [5 x (5 x) + 5 x \cdot 1 + 1 (5 x + 1)]$

Step 1.2.3

Apply the distributive property.

$\frac{d}{d x} [5 x (5 x) + 5 x \cdot 1 + 1 (5 x) + 1 \cdot 1]$

Step 1.3

Simplify and combine like terms.

Tap for more steps...

Step 1.3.1

Simplify each term.

Tap for more steps...

Step 1.3.1.1

Rewrite using the commutative property of multiplication.

$\frac{d}{d x} [5 \cdot 5 x \cdot x + 5 x \cdot 1 + 1 (5 x) + 1 \cdot 1]$

Step 1.3.1.2

Multiply $x$ by $x$ by adding the exponents.

Tap for more steps...

Step 1.3.1.2.1

Move $x$ .

$\frac{d}{d x} [5 \cdot 5 (x \cdot x) + 5 x \cdot 1 + 1 (5 x) + 1 \cdot 1]$

Step 1.3.1.2.2

Multiply $x$ by $x$ .

$\frac{d}{d x} [5 \cdot 5 x^{2} + 5 x \cdot 1 + 1 (5 x) + 1 \cdot 1]$

Step 1.3.1.3

Multiply $5$ by $5$ .

$\frac{d}{d x} [25 x^{2} + 5 x \cdot 1 + 1 (5 x) + 1 \cdot 1]$

Step 1.3.1.4

Multiply $5$ by $1$ .

$\frac{d}{d x} [25 x^{2} + 5 x + 1 (5 x) + 1 \cdot 1]$

Step 1.3.1.5

Multiply $5 x$ by $1$ .

$\frac{d}{d x} [25 x^{2} + 5 x + 5 x + 1 \cdot 1]$

Step 1.3.1.6

Multiply $1$ by $1$ .

$\frac{d}{d x} [25 x^{2} + 5 x + 5 x + 1]$

Step 1.3.2

Add $5 x$ and $5 x$ .

$\frac{d}{d x} [25 x^{2} + 10 x + 1]$

Step 1.4

By the Sum Rule, the derivative of $25 x^{2} + 10 x + 1$ with respect to $x$ is $\frac{d}{d x} [25 x^{2}] + \frac{d}{d x} [10 x] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [25 x^{2}] + \frac{d}{d x} [10 x] + \frac{d}{d x} [1]$

Step 1.5

Since $25$ is constant with respect to $x$ , the derivative of $25 x^{2}$ with respect to $x$ is $25 \frac{d}{d x} [x^{2}]$ .

$25 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [10 x] + \frac{d}{d x} [1]$

Step 1.6

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

$25 (2 x) + \frac{d}{d x} [10 x] + \frac{d}{d x} [1]$

Step 1.7

Multiply $2$ by $25$ .

$50 x + \frac{d}{d x} [10 x] + \frac{d}{d x} [1]$

Step 1.8

Since $10$ is constant with respect to $x$ , the derivative of $10 x$ with respect to $x$ is $10 \frac{d}{d x} [x]$ .

$50 x + 10 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 1.9

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

$50 x + 10 \cdot 1 + \frac{d}{d x} [1]$

Step 1.10

Multiply $10$ by $1$ .

$50 x + 10 + \frac{d}{d x} [1]$

Step 1.11

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

$50 x + 10 + 0$

Step 1.12

Add $50 x + 10$ and $0$ .

$50 x + 10$

Step 1.13

Evaluate the derivative at $x = 0$ .

$50 (0) + 10$

Step 1.14

Simplify.

Tap for more steps...

Step 1.14.1

Multiply $50$ by $0$ .

$0 + 10$

Step 1.14.2

Add $0$ and $10$ .

$10$

Step 2

Plug the slope and point values into the point-slope formula and solve for $y$ .

Tap for more steps...

Step 2.1

Use the slope $10$ and a given point $(0, 1)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (1) = 10 \cdot (x - (0))$

Step 2.2

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

$y - 1 = 10 \cdot (x + 0)$

Step 2.3

Solve for $y$ .

Tap for more steps...

Step 2.3.1

Add $x$ and $0$ .

$y - 1 = 10 x$

Step 2.3.2

Add $1$ to both sides of the equation.

$y = 10 x + 1$

Step 3