Calculus Examples

$z (t) = \frac{t^{2} + 5 t + 2}{t + 9}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d t} [\frac{f (t)}{g (t)}]$ is $\frac{g (t) \frac{d}{d t} [f (t)] - f (t) \frac{d}{d t} [g (t)]}{{g (t)}^{2}}$ where $f (t) = t^{2} + 5 t + 2$ and $g (t) = t + 9$ .

$\frac{(t + 9) \frac{d}{d t} [t^{2} + 5 t + 2] - (t^{2} + 5 t + 2) \frac{d}{d t} [t + 9]}{{(t + 9)}^{2}}$

Step 2

Differentiate.

Tap for more steps...

Step 2.1

By the Sum Rule, the derivative of $t^{2} + 5 t + 2$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [5 t] + \frac{d}{d t} [2]$ .

$\frac{(t + 9) (\frac{d}{d t} [t^{2}] + \frac{d}{d t} [5 t] + \frac{d}{d t} [2]) - (t^{2} + 5 t + 2) \frac{d}{d t} [t + 9]}{{(t + 9)}^{2}}$

Step 2.2

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

$\frac{(t + 9) (2 t + \frac{d}{d t} [5 t] + \frac{d}{d t} [2]) - (t^{2} + 5 t + 2) \frac{d}{d t} [t + 9]}{{(t + 9)}^{2}}$

Step 2.3

Since $5$ is constant with respect to $t$ , the derivative of $5 t$ with respect to $t$ is $5 \frac{d}{d t} [t]$ .

$\frac{(t + 9) (2 t + 5 \frac{d}{d t} [t] + \frac{d}{d t} [2]) - (t^{2} + 5 t + 2) \frac{d}{d t} [t + 9]}{{(t + 9)}^{2}}$

Step 2.4

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

$\frac{(t + 9) (2 t + 5 \cdot 1 + \frac{d}{d t} [2]) - (t^{2} + 5 t + 2) \frac{d}{d t} [t + 9]}{{(t + 9)}^{2}}$

Step 2.5

Multiply $5$ by $1$ .

$\frac{(t + 9) (2 t + 5 + \frac{d}{d t} [2]) - (t^{2} + 5 t + 2) \frac{d}{d t} [t + 9]}{{(t + 9)}^{2}}$

Step 2.6

Since $2$ is constant with respect to $t$ , the derivative of $2$ with respect to $t$ is $0$ .

$\frac{(t + 9) (2 t + 5 + 0) - (t^{2} + 5 t + 2) \frac{d}{d t} [t + 9]}{{(t + 9)}^{2}}$

Step 2.7

Add $2 t + 5$ and $0$ .

$\frac{(t + 9) (2 t + 5) - (t^{2} + 5 t + 2) \frac{d}{d t} [t + 9]}{{(t + 9)}^{2}}$

Step 2.8

By the Sum Rule, the derivative of $t + 9$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [9]$ .

$\frac{(t + 9) (2 t + 5) - (t^{2} + 5 t + 2) (\frac{d}{d t} [t] + \frac{d}{d t} [9])}{{(t + 9)}^{2}}$

Step 2.9

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

$\frac{(t + 9) (2 t + 5) - (t^{2} + 5 t + 2) (1 + \frac{d}{d t} [9])}{{(t + 9)}^{2}}$

Step 2.10

Since $9$ is constant with respect to $t$ , the derivative of $9$ with respect to $t$ is $0$ .

$\frac{(t + 9) (2 t + 5) - (t^{2} + 5 t + 2) (1 + 0)}{{(t + 9)}^{2}}$

Step 2.11

Simplify the expression.

Tap for more steps...

Step 2.11.1

Add $1$ and $0$ .

$\frac{(t + 9) (2 t + 5) - (t^{2} + 5 t + 2) \cdot 1}{{(t + 9)}^{2}}$

Step 2.11.2

Multiply $- 1$ by $1$ .

$\frac{(t + 9) (2 t + 5) - (t^{2} + 5 t + 2)}{{(t + 9)}^{2}}$

Step 3

Simplify.

Tap for more steps...

Step 3.1

Apply the distributive property.

$\frac{(t + 9) (2 t + 5) - t^{2} - (5 t) - 1 \cdot 2}{{(t + 9)}^{2}}$

Step 3.2

Simplify the numerator.

Tap for more steps...

Step 3.2.1

Simplify each term.

Tap for more steps...

Step 3.2.1.1

Expand $(t + 9) (2 t + 5)$ using the FOIL Method.

Tap for more steps...

Step 3.2.1.1.1

Apply the distributive property.

$\frac{t (2 t + 5) + 9 (2 t + 5) - t^{2} - (5 t) - 1 \cdot 2}{{(t + 9)}^{2}}$

Step 3.2.1.1.2

Apply the distributive property.

$\frac{t (2 t) + t \cdot 5 + 9 (2 t + 5) - t^{2} - (5 t) - 1 \cdot 2}{{(t + 9)}^{2}}$

Step 3.2.1.1.3

Apply the distributive property.

$\frac{t (2 t) + t \cdot 5 + 9 (2 t) + 9 \cdot 5 - t^{2} - (5 t) - 1 \cdot 2}{{(t + 9)}^{2}}$

Step 3.2.1.2

Simplify and combine like terms.

Tap for more steps...

Step 3.2.1.2.1

Simplify each term.

Tap for more steps...

Step 3.2.1.2.1.1

Rewrite using the commutative property of multiplication.

$\frac{2 t \cdot t + t \cdot 5 + 9 (2 t) + 9 \cdot 5 - t^{2} - (5 t) - 1 \cdot 2}{{(t + 9)}^{2}}$

Step 3.2.1.2.1.2

Multiply $t$ by $t$ by adding the exponents.

Tap for more steps...

Step 3.2.1.2.1.2.1

Move $t$ .

$\frac{2 (t \cdot t) + t \cdot 5 + 9 (2 t) + 9 \cdot 5 - t^{2} - (5 t) - 1 \cdot 2}{{(t + 9)}^{2}}$

Step 3.2.1.2.1.2.2

Multiply $t$ by $t$ .

$\frac{2 t^{2} + t \cdot 5 + 9 (2 t) + 9 \cdot 5 - t^{2} - (5 t) - 1 \cdot 2}{{(t + 9)}^{2}}$

Step 3.2.1.2.1.3

Move $5$ to the left of $t$ .

$\frac{2 t^{2} + 5 \cdot t + 9 (2 t) + 9 \cdot 5 - t^{2} - (5 t) - 1 \cdot 2}{{(t + 9)}^{2}}$

Step 3.2.1.2.1.4

Multiply $2$ by $9$ .

$\frac{2 t^{2} + 5 t + 18 t + 9 \cdot 5 - t^{2} - (5 t) - 1 \cdot 2}{{(t + 9)}^{2}}$

Step 3.2.1.2.1.5

Multiply $9$ by $5$ .

$\frac{2 t^{2} + 5 t + 18 t + 45 - t^{2} - (5 t) - 1 \cdot 2}{{(t + 9)}^{2}}$

Step 3.2.1.2.2

Add $5 t$ and $18 t$ .

$\frac{2 t^{2} + 23 t + 45 - t^{2} - (5 t) - 1 \cdot 2}{{(t + 9)}^{2}}$

Step 3.2.1.3

Multiply $5$ by $- 1$ .

$\frac{2 t^{2} + 23 t + 45 - t^{2} - 5 t - 1 \cdot 2}{{(t + 9)}^{2}}$

Step 3.2.1.4

Multiply $- 1$ by $2$ .

$\frac{2 t^{2} + 23 t + 45 - t^{2} - 5 t - 2}{{(t + 9)}^{2}}$

Step 3.2.2

Subtract $t^{2}$ from $2 t^{2}$ .

$\frac{t^{2} + 23 t + 45 - 5 t - 2}{{(t + 9)}^{2}}$

Step 3.2.3

Subtract $5 t$ from $23 t$ .

$\frac{t^{2} + 18 t + 45 - 2}{{(t + 9)}^{2}}$

Step 3.2.4

Subtract $2$ from $45$ .

$\frac{t^{2} + 18 t + 43}{{(t + 9)}^{2}}$