Calculus Examples

$\frac{80 \sin (0.6 t)}{t} + 15$

Step 1

By the Sum Rule, the derivative of $\frac{80 \sin (0.6 t)}{t} + 15$ with respect to $t$ is $\frac{d}{d t} [\frac{80 \sin (0.6 t)}{t}] + \frac{d}{d t} [15]$ .

$\frac{d}{d t} [\frac{80 \sin (0.6 t)}{t}] + \frac{d}{d t} [15]$

Step 2

Evaluate $\frac{d}{d t} [\frac{80 \sin (0.6 t)}{t}]$ .

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Step 2.1

Since $80$ is constant with respect to $t$ , the derivative of $\frac{80 \sin (0.6 t)}{t}$ with respect to $t$ is $80 \frac{d}{d t} [\frac{\sin (0.6 t)}{t}]$ .

$80 \frac{d}{d t} [\frac{\sin (0.6 t)}{t}] + \frac{d}{d t} [15]$

Step 2.2

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 $f (t) = \sin (0.6 t)$ and $g (t) = t$ .

$80 \frac{t \frac{d}{d t} [\sin (0.6 t)] - \sin (0.6 t) \frac{d}{d t} [t]}{t^{2}} + \frac{d}{d t} [15]$

Step 2.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \sin (t)$ and $g (t) = 0.6 t$ .

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Step 2.3.1

To apply the Chain Rule, set $u$ as $0.6 t$ .

$80 \frac{t (\frac{d}{d u} [\sin (u)] \frac{d}{d t} [0.6 t]) - \sin (0.6 t) \frac{d}{d t} [t]}{t^{2}} + \frac{d}{d t} [15]$

Step 2.3.2

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

$80 \frac{t (\cos (u) \frac{d}{d t} [0.6 t]) - \sin (0.6 t) \frac{d}{d t} [t]}{t^{2}} + \frac{d}{d t} [15]$

Step 2.3.3

Replace all occurrences of $u$ with $0.6 t$ .

$80 \frac{t (\cos (0.6 t) \frac{d}{d t} [0.6 t]) - \sin (0.6 t) \frac{d}{d t} [t]}{t^{2}} + \frac{d}{d t} [15]$

Step 2.4

Since $0.6$ is constant with respect to $t$ , the derivative of $0.6 t$ with respect to $t$ is $0.6 \frac{d}{d t} [t]$ .

$80 \frac{t (\cos (0.6 t) (0.6 \frac{d}{d t} [t])) - \sin (0.6 t) \frac{d}{d t} [t]}{t^{2}} + \frac{d}{d t} [15]$

Step 2.5

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

$80 \frac{t (\cos (0.6 t) (0.6 \cdot 1)) - \sin (0.6 t) \frac{d}{d t} [t]}{t^{2}} + \frac{d}{d t} [15]$

Step 2.6

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

$80 \frac{t (\cos (0.6 t) (0.6 \cdot 1)) - \sin (0.6 t) \cdot 1}{t^{2}} + \frac{d}{d t} [15]$

Step 2.7

Multiply $0.6$ by $1$ .

$80 \frac{t (\cos (0.6 t) \cdot 0.6) - \sin (0.6 t) \cdot 1}{t^{2}} + \frac{d}{d t} [15]$

Step 2.8

Move $0.6$ to the left of $\cos (0.6 t)$ .

$80 \frac{t (0.6 \cdot \cos (0.6 t)) - \sin (0.6 t) \cdot 1}{t^{2}} + \frac{d}{d t} [15]$

Step 2.9

Multiply $- 1$ by $1$ .

$80 \frac{t (0.6 \cos (0.6 t)) - \sin (0.6 t)}{t^{2}} + \frac{d}{d t} [15]$

Step 2.10

Combine $80$ and $\frac{t (0.6 \cos (0.6 t)) - \sin (0.6 t)}{t^{2}}$ .

$\frac{80 (t (0.6 \cos (0.6 t)) - \sin (0.6 t))}{t^{2}} + \frac{d}{d t} [15]$

Step 3

Since $15$ is constant with respect to $t$ , the derivative of $15$ with respect to $t$ is $0$ .

$\frac{80 (t (0.6 \cos (0.6 t)) - \sin (0.6 t))}{t^{2}} + 0$

Step 4

Simplify.

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Step 4.1

Apply the distributive property.

$\frac{80 (t (0.6 \cos (0.6 t))) + 80 (- \sin (0.6 t))}{t^{2}} + 0$

Step 4.2

Combine terms.

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Step 4.2.1

Multiply $0.6$ by $80$ .

$\frac{48 (t (\cos (0.6 t))) + 80 (- \sin (0.6 t))}{t^{2}} + 0$

Step 4.2.2

Multiply $- 1$ by $80$ .

$\frac{48 t \cos (0.6 t) - 80 \sin (0.6 t)}{t^{2}} + 0$

Step 4.2.3

Add $\frac{48 t \cos (0.6 t) - 80 \sin (0.6 t)}{t^{2}}$ and $0$ .

$\frac{48 t \cos (0.6 t) - 80 \sin (0.6 t)}{t^{2}}$

Step 4.3

Factor $16$ out of $48 t \cos (0.6 t) - 80 \sin (0.6 t)$ .

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Step 4.3.1

Factor $16$ out of $48 t \cos (0.6 t)$ .

$\frac{16 (3 t \cos (0.6 t)) - 80 \sin (0.6 t)}{t^{2}}$

Step 4.3.2

Factor $16$ out of $- 80 \sin (0.6 t)$ .

$\frac{16 (3 t \cos (0.6 t)) + 16 (- 5 \sin (0.6 t))}{t^{2}}$

Step 4.3.3

Factor $16$ out of $16 (3 t \cos (0.6 t)) + 16 (- 5 \sin (0.6 t))$ .

$\frac{16 (3 t \cos (0.6 t) - 5 \sin (0.6 t))}{t^{2}}$