Calculus Examples

$A \sin (t) + \cos (k t)$

Step 1

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

$\frac{d}{d t} [A \sin (t)] + \frac{d}{d t} [\cos (k t)]$

Step 2

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

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Step 2.1

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

$A \frac{d}{d t} [\sin (t)] + \frac{d}{d t} [\cos (k t)]$

Step 2.2

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

$A \cos (t) + \frac{d}{d t} [\cos (k t)]$

Step 3

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

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Step 3.1

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 $g (t) = k t$ .

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Step 3.1.1

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

$A \cos (t) + \frac{d}{d u} [\cos (u)] \frac{d}{d t} [k t]$

Step 3.1.2

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

$A \cos (t) - \sin (u) \frac{d}{d t} [k t]$

Step 3.1.3

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

$A \cos (t) - \sin (k t) \frac{d}{d t} [k t]$

Step 3.2

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

$A \cos (t) - \sin (k t) (k \frac{d}{d t} [t])$

Step 3.3

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

$A \cos (t) - \sin (k t) (k \cdot 1)$

Step 3.4

Multiply $k$ by $1$ .

$A \cos (t) - \sin (k t) k$

Step 4

Reorder terms.

$A \cos (t) - k \sin (k t)$