Calculus Examples

$y = 5 + 4.9 \cos (\frac{π}{6} t)$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d t} (y) = \frac{d}{d t} (5 + 4.9 \cos (\frac{π}{6} t))$

Step 2

The derivative of $y$ with respect to $t$ is $y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

Tap for more steps...

Step 3.1

Differentiate.

Tap for more steps...

Step 3.1.1

By the Sum Rule, the derivative of $5 + 4.9 \cos (\frac{π}{6} t)$ with respect to $t$ is $\frac{d}{d t} [5] + \frac{d}{d t} [4.9 \cos (\frac{π}{6} t)]$ .

$\frac{d}{d t} [5] + \frac{d}{d t} [4.9 \cos (\frac{π}{6} t)]$

Step 3.1.2

Since $5$ is constant with respect to $t$ , the derivative of $5$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [4.9 \cos (\frac{π}{6} t)]$

Step 3.2

Evaluate $\frac{d}{d t} [4.9 \cos (\frac{π}{6} t)]$ .

Tap for more steps...

Step 3.2.1

Combine $\frac{π}{6}$ and $t$ .

$0 + \frac{d}{d t} [4.9 \cos (\frac{π t}{6})]$

Step 3.2.2

Since $4.9$ is constant with respect to $t$ , the derivative of $4.9 \cos (\frac{π t}{6})$ with respect to $t$ is $4.9 \frac{d}{d t} [\cos (\frac{π t}{6})]$ .

$0 + 4.9 \frac{d}{d t} [\cos (\frac{π t}{6})]$

Step 3.2.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \cos (t)$ and $g (t) = \frac{π t}{6}$ .

Tap for more steps...

Step 3.2.3.1

To apply the Chain Rule, set $u$ as $\frac{π t}{6}$ .

$0 + 4.9 (\frac{d}{d u} [\cos (u)] \frac{d}{d t} [\frac{π t}{6}])$

Step 3.2.3.2

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

$0 + 4.9 (- \sin (u) \frac{d}{d t} [\frac{π t}{6}])$

Step 3.2.3.3

Replace all occurrences of $u$ with $\frac{π t}{6}$ .

$0 + 4.9 (- \sin (\frac{π t}{6}) \frac{d}{d t} [\frac{π t}{6}])$

Step 3.2.4

Since $\frac{π}{6}$ is constant with respect to $t$ , the derivative of $\frac{π t}{6}$ with respect to $t$ is $\frac{π}{6} \frac{d}{d t} [t]$ .

$0 + 4.9 (- \sin (\frac{π t}{6}) (\frac{π}{6} \frac{d}{d t} [t]))$

Step 3.2.5

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

$0 + 4.9 (- \sin (\frac{π t}{6}) (\frac{π}{6} \cdot 1))$

Step 3.2.6

Multiply $\frac{π}{6}$ by $1$ .

$0 + 4.9 (- \sin (\frac{π t}{6}) \frac{π}{6})$

Step 3.2.7

Combine $\frac{π}{6}$ and $\sin (\frac{π t}{6})$ .

$0 + 4.9 (- \frac{π \sin (\frac{π t}{6})}{6})$

Step 3.2.8

Multiply $- 1$ by $4.9$ .

$0 - 4.9 \frac{π \sin (\frac{π t}{6})}{6}$

Step 3.2.9

Combine $- 4.9$ and $\frac{π \sin (\frac{π t}{6})}{6}$ .

$0 + \frac{- 4.9 (π \sin (\frac{π t}{6}))}{6}$

Step 3.2.10

Multiply $π$ by $- 4.9$ .

$0 + \frac{- 15.393804 \sin (\frac{π t}{6})}{6}$

Step 3.2.11

Move the negative in front of the fraction.

$0 - \frac{15.393804 \sin (\frac{π t}{6})}{6}$

Step 3.3

Subtract $\frac{15.393804 \sin (\frac{π t}{6})}{6}$ from $0$ .

$- \frac{15.393804 \sin (\frac{π t}{6})}{6}$

Step 4

Reform the equation by setting the left side equal to the right side.

$y' = - \frac{15.393804 \sin (\frac{π t}{6})}{6}$

Step 5

Replace $y'$ with $\frac{d y}{d t}$ .

$\frac{d y}{d t} = - \frac{15.393804 \sin (\frac{π t}{6})}{6}$