Calculus Examples

$g (t) = (\frac{1 + \sin (3 t)}{3 - 2 t}) - 1$

Step 1

By the Sum Rule, the derivative of $(\frac{1 + \sin (3 t)}{3 - 2 t}) - 1$ with respect to $t$ is $\frac{d}{d t} [\frac{1 + \sin (3 t)}{3 - 2 t}] + \frac{d}{d t} [- 1]$ .

$\frac{d}{d t} [\frac{1 + \sin (3 t)}{3 - 2 t}] + \frac{d}{d t} [- 1]$

Step 2

Evaluate $\frac{d}{d t} [\frac{1 + \sin (3 t)}{3 - 2 t}]$ .

Tap for more steps...

Step 2.1

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 $f (t) = 1 + \sin (3 t)$ and $g (t) = 3 - 2 t$ .

$\frac{(3 - 2 t) \frac{d}{d t} [1 + \sin (3 t)] - (1 + \sin (3 t)) \frac{d}{d t} [3 - 2 t]}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.2

By the Sum Rule, the derivative of $1 + \sin (3 t)$ with respect to $t$ is $\frac{d}{d t} [1] + \frac{d}{d t} [\sin (3 t)]$ .

$\frac{(3 - 2 t) (\frac{d}{d t} [1] + \frac{d}{d t} [\sin (3 t)]) - (1 + \sin (3 t)) \frac{d}{d t} [3 - 2 t]}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.3

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

$\frac{(3 - 2 t) (0 + \frac{d}{d t} [\sin (3 t)]) - (1 + \sin (3 t)) \frac{d}{d t} [3 - 2 t]}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.4

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \sin (t)$ and $g (t) = 3 t$ .

Tap for more steps...

Step 2.4.1

To apply the Chain Rule, set $u$ as $3 t$ .

$\frac{(3 - 2 t) (0 + \frac{d}{d u} [\sin (u)] \frac{d}{d t} [3 t]) - (1 + \sin (3 t)) \frac{d}{d t} [3 - 2 t]}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.4.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$\frac{(3 - 2 t) (0 + \cos (u) \frac{d}{d t} [3 t]) - (1 + \sin (3 t)) \frac{d}{d t} [3 - 2 t]}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.4.3

Replace all occurrences of $u$ with $3 t$ .

$\frac{(3 - 2 t) (0 + \cos (3 t) \frac{d}{d t} [3 t]) - (1 + \sin (3 t)) \frac{d}{d t} [3 - 2 t]}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.5

Since $3$ is constant with respect to $t$ , the derivative of $3 t$ with respect to $t$ is $3 \frac{d}{d t} [t]$ .

$\frac{(3 - 2 t) (0 + \cos (3 t) (3 \frac{d}{d t} [t])) - (1 + \sin (3 t)) \frac{d}{d t} [3 - 2 t]}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.6

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

$\frac{(3 - 2 t) (0 + \cos (3 t) (3 \cdot 1)) - (1 + \sin (3 t)) \frac{d}{d t} [3 - 2 t]}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.7

By the Sum Rule, the derivative of $3 - 2 t$ with respect to $t$ is $\frac{d}{d t} [3] + \frac{d}{d t} [- 2 t]$ .

$\frac{(3 - 2 t) (0 + \cos (3 t) (3 \cdot 1)) - (1 + \sin (3 t)) (\frac{d}{d t} [3] + \frac{d}{d t} [- 2 t])}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.8

Since $3$ is constant with respect to $t$ , the derivative of $3$ with respect to $t$ is $0$ .

$\frac{(3 - 2 t) (0 + \cos (3 t) (3 \cdot 1)) - (1 + \sin (3 t)) (0 + \frac{d}{d t} [- 2 t])}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.9

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

$\frac{(3 - 2 t) (0 + \cos (3 t) (3 \cdot 1)) - (1 + \sin (3 t)) (0 - 2 \frac{d}{d t} [t])}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.10

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

$\frac{(3 - 2 t) (0 + \cos (3 t) (3 \cdot 1)) - (1 + \sin (3 t)) (0 - 2 \cdot 1)}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.11

Multiply $3$ by $1$ .

$\frac{(3 - 2 t) (0 + \cos (3 t) \cdot 3) - (1 + \sin (3 t)) (0 - 2 \cdot 1)}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.12

Move $3$ to the left of $\cos (3 t)$ .

$\frac{(3 - 2 t) (0 + 3 \cdot \cos (3 t)) - (1 + \sin (3 t)) (0 - 2 \cdot 1)}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.13

Add $0$ and $3 \cos (3 t)$ .

$\frac{(3 - 2 t) (3 \cos (3 t)) - (1 + \sin (3 t)) (0 - 2 \cdot 1)}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.14

Move $3$ to the left of $3 - 2 t$ .

$\frac{3 \cdot (3 - 2 t) \cos (3 t) - (1 + \sin (3 t)) (0 - 2 \cdot 1)}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.15

Multiply $- 2$ by $1$ .

$\frac{3 (3 - 2 t) \cos (3 t) - (1 + \sin (3 t)) (0 - 2)}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.16

Subtract $2$ from $0$ .

$\frac{3 (3 - 2 t) \cos (3 t) - (1 + \sin (3 t)) \cdot - 2}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 2.17

Multiply $- 2$ by $- 1$ .

$\frac{3 (3 - 2 t) \cos (3 t) + 2 (1 + \sin (3 t))}{{(3 - 2 t)}^{2}} + \frac{d}{d t} [- 1]$

Step 3

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

$\frac{3 (3 - 2 t) \cos (3 t) + 2 (1 + \sin (3 t))}{{(3 - 2 t)}^{2}} + 0$

Step 4

Simplify.

Tap for more steps...

Step 4.1

Apply the distributive property.

$\frac{(3 \cdot 3 + 3 (- 2 t)) \cos (3 t) + 2 (1 + \sin (3 t))}{{(3 - 2 t)}^{2}} + 0$

Step 4.2

Apply the distributive property.

$\frac{3 \cdot 3 \cos (3 t) + 3 (- 2 t) \cos (3 t) + 2 (1 + \sin (3 t))}{{(3 - 2 t)}^{2}} + 0$

Step 4.3

Apply the distributive property.

$\frac{3 \cdot 3 \cos (3 t) + 3 (- 2 t) \cos (3 t) + 2 \cdot 1 + 2 \sin (3 t)}{{(3 - 2 t)}^{2}} + 0$

Step 4.4

Combine terms.

Tap for more steps...

Step 4.4.1

Multiply $3$ by $3$ .

$\frac{9 \cos (3 t) + 3 (- 2 t) \cos (3 t) + 2 \cdot 1 + 2 \sin (3 t)}{{(3 - 2 t)}^{2}} + 0$

Step 4.4.2

Multiply $- 2$ by $3$ .

$\frac{9 \cos (3 t) - 6 t \cos (3 t) + 2 \cdot 1 + 2 \sin (3 t)}{{(3 - 2 t)}^{2}} + 0$

Step 4.4.3

Multiply $2$ by $1$ .

$\frac{9 \cos (3 t) - 6 t \cos (3 t) + 2 + 2 \sin (3 t)}{{(3 - 2 t)}^{2}} + 0$

Step 4.4.4

Add $\frac{9 \cos (3 t) - 6 t \cos (3 t) + 2 + 2 \sin (3 t)}{{(3 - 2 t)}^{2}}$ and $0$ .

$\frac{9 \cos (3 t) - 6 t \cos (3 t) + 2 + 2 \sin (3 t)}{{(3 - 2 t)}^{2}}$

Step 4.5

Reorder terms.

$\frac{- 6 t \cos (3 t) + 2 + 9 \cos (3 t) + 2 \sin (3 t)}{{(- 2 t + 3)}^{2}}$