Calculus Examples

$f (x) = \arcsin (t) + 3 t^{6} + 6$

Step 1

By the Sum Rule, the derivative of $\arcsin (t) + 3 t^{6} + 6$ with respect to $t$ is $\frac{d}{d t} [\arcsin (t)] + \frac{d}{d t} [3 t^{6}] + \frac{d}{d t} [6]$ .

$\frac{d}{d t} [\arcsin (t)] + \frac{d}{d t} [3 t^{6}] + \frac{d}{d t} [6]$

Step 2

The derivative of $\arcsin (t)$ with respect to $t$ is $\frac{1}{\sqrt{1 - t^{2}}}$ .

$\frac{1}{\sqrt{1 - t^{2}}} + \frac{d}{d t} [3 t^{6}] + \frac{d}{d t} [6]$

Step 3

Evaluate $\frac{d}{d t} [3 t^{6}]$ .

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Step 3.1

Since $3$ is constant with respect to $t$ , the derivative of $3 t^{6}$ with respect to $t$ is $3 \frac{d}{d t} [t^{6}]$ .

$\frac{1}{\sqrt{1 - t^{2}}} + 3 \frac{d}{d t} [t^{6}] + \frac{d}{d t} [6]$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 6$ .

$\frac{1}{\sqrt{1 - t^{2}}} + 3 (6 t^{5}) + \frac{d}{d t} [6]$

Step 3.3

Multiply $6$ by $3$ .

$\frac{1}{\sqrt{1 - t^{2}}} + 18 t^{5} + \frac{d}{d t} [6]$

Step 4

Since $6$ is constant with respect to $t$ , the derivative of $6$ with respect to $t$ is $0$ .

$\frac{1}{\sqrt{1 - t^{2}}} + 18 t^{5} + 0$

Step 5

Simplify.

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Step 5.1

Add $\frac{1}{\sqrt{1 - t^{2}}} + 18 t^{5}$ and $0$ .

$\frac{1}{\sqrt{1 - t^{2}}} + 18 t^{5}$

Step 5.2

Reorder terms.

$18 t^{5} + \frac{1}{\sqrt{1 - t^{2}}}$