Calculus Examples

$h (t) = 12 \csc (t) + t \cot (t)$

Step 1

By the Sum Rule, the derivative of $12 \csc (t) + t \cot (t)$ with respect to $t$ is $\frac{d}{d t} [12 \csc (t)] + \frac{d}{d t} [t \cot (t)]$ .

$\frac{d}{d t} [12 \csc (t)] + \frac{d}{d t} [t \cot (t)]$

Step 2

Evaluate $\frac{d}{d t} [12 \csc (t)]$ .

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Step 2.1

Since $12$ is constant with respect to $t$ , the derivative of $12 \csc (t)$ with respect to $t$ is $12 \frac{d}{d t} [\csc (t)]$ .

$12 \frac{d}{d t} [\csc (t)] + \frac{d}{d t} [t \cot (t)]$

Step 2.2

The derivative of $\csc (t)$ with respect to $t$ is $- \csc (t) \cot (t)$ .

$12 (- \csc (t) \cot (t)) + \frac{d}{d t} [t \cot (t)]$

Step 2.3

Multiply $- 1$ by $12$ .

$- 12 \csc (t) \cot (t) + \frac{d}{d t} [t \cot (t)]$

Step 3

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

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Step 3.1

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = t$ and $g (t) = \cot (t)$ .

$- 12 \csc (t) \cot (t) + t \frac{d}{d t} [\cot (t)] + \cot (t) \frac{d}{d t} [t]$

Step 3.2

The derivative of $\cot (t)$ with respect to $t$ is $- \csc^{2} (t)$ .

$- 12 \csc (t) \cot (t) + t (- \csc^{2} (t)) + \cot (t) \frac{d}{d t} [t]$

Step 3.3

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

$- 12 \csc (t) \cot (t) + t (- \csc^{2} (t)) + \cot (t) \cdot 1$

Step 3.4

Multiply $\cot (t)$ by $1$ .

$- 12 \csc (t) \cot (t) + t (- \csc^{2} (t)) + \cot (t)$

Step 4

Reorder terms.

$- 12 \cot (t) \csc (t) - t \csc^{2} (t) + \cot (t)$