Calculus Examples

$s = 5 - 3 \cos (t)$

Step 1

Find the first derivative.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of $5 - 3 \cos (t)$ with respect to $t$ is $\frac{d}{d t} [5] + \frac{d}{d t} [- 3 \cos (t)]$ .

$\frac{d}{d t} [5] + \frac{d}{d t} [- 3 \cos (t)]$

Step 1.1.2

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

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

Step 1.2

Evaluate $\frac{d}{d t} [- 3 \cos (t)]$ .

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Step 1.2.1

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

$0 - 3 \frac{d}{d t} [\cos (t)]$

Step 1.2.2

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

$0 - 3 (- \sin (t))$

Step 1.2.3

Multiply $- 1$ by $- 3$ .

$0 + 3 \sin (t)$

Step 1.3

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

$f' (t) = 3 \sin (t)$

Step 2

Find the second derivative.

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Step 2.1

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

$3 \frac{d}{d t} [\sin (t)]$

Step 2.2

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

$f'' (t) = 3 \cos (t)$