Calculus Examples

$g (t) = \sin^{2} (3 t) + \arcsin (3 t)$

Step 1

By the Sum Rule, the derivative of $\sin^{2} (3 t) + \arcsin (3 t)$ with respect to $t$ is $\frac{d}{d t} [\sin^{2} (3 t)] + \frac{d}{d t} [\arcsin (3 t)]$ .

$\frac{d}{d t} [\sin^{2} (3 t)] + \frac{d}{d t} [\arcsin (3 t)]$

Step 2

Evaluate $\frac{d}{d t} [\sin^{2} (3 t)]$ .

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Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{2}$ and $g (t) = \sin (3 t)$ .

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Step 2.1.1

To apply the Chain Rule, set $u_{1}$ as $\sin (3 t)$ .

$\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d t} [\sin (3 t)] + \frac{d}{d t} [\arcsin (3 t)]$

Step 2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 2$ .

$2 u_{1} \frac{d}{d t} [\sin (3 t)] + \frac{d}{d t} [\arcsin (3 t)]$

Step 2.1.3

Replace all occurrences of $u_{1}$ with $\sin (3 t)$ .

$2 \sin (3 t) \frac{d}{d t} [\sin (3 t)] + \frac{d}{d t} [\arcsin (3 t)]$

Step 2.2

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 $g (t) = 3 t$ .

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Step 2.2.1

To apply the Chain Rule, set $u_{2}$ as $3 t$ .

$2 \sin (3 t) (\frac{d}{d u_{2}} [\sin (u_{2})] \frac{d}{d t} [3 t]) + \frac{d}{d t} [\arcsin (3 t)]$

Step 2.2.2

The derivative of $\sin (u_{2})$ with respect to $u_{2}$ is $\cos (u_{2})$ .

$2 \sin (3 t) (\cos (u_{2}) \frac{d}{d t} [3 t]) + \frac{d}{d t} [\arcsin (3 t)]$

Step 2.2.3

Replace all occurrences of $u_{2}$ with $3 t$ .

$2 \sin (3 t) (\cos (3 t) \frac{d}{d t} [3 t]) + \frac{d}{d t} [\arcsin (3 t)]$

Step 2.3

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

$2 \sin (3 t) (\cos (3 t) (3 \frac{d}{d t} [t])) + \frac{d}{d t} [\arcsin (3 t)]$

Step 2.4

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

$2 \sin (3 t) (\cos (3 t) (3 \cdot 1)) + \frac{d}{d t} [\arcsin (3 t)]$

Step 2.5

Multiply $3$ by $1$ .

$2 \sin (3 t) (\cos (3 t) \cdot 3) + \frac{d}{d t} [\arcsin (3 t)]$

Step 2.6

Move $3$ to the left of $\cos (3 t)$ .

$2 \sin (3 t) (3 \cdot \cos (3 t)) + \frac{d}{d t} [\arcsin (3 t)]$

Step 2.7

Multiply $3$ by $2$ .

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

Step 3

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

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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 $f (t) = \arcsin (t)$ and $g (t) = 3 t$ .

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Step 3.1.1

To apply the Chain Rule, set $u_{3}$ as $3 t$ .

$6 \sin (3 t) \cos (3 t) + \frac{d}{d u_{3}} [\arcsin (u_{3})] \frac{d}{d t} [3 t]$

Step 3.1.2

The derivative of $\arcsin (u_{3})$ with respect to $u_{3}$ is $\frac{1}{\sqrt{1 - {u_{3}}^{2}}}$ .

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

Step 3.1.3

Replace all occurrences of $u_{3}$ with $3 t$ .

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

Step 3.2

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

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

Step 3.3

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

$6 \sin (3 t) \cos (3 t) + \frac{1}{\sqrt{1 - {(3 t)}^{2}}} (3 \cdot 1)$

Step 3.4

Factor $3$ out of $3 t$ .

$6 \sin (3 t) \cos (3 t) + \frac{1}{\sqrt{1 - {(3 (t))}^{2}}} (3 \cdot 1)$

Step 3.5

Apply the product rule to $3 (t)$ .

$6 \sin (3 t) \cos (3 t) + \frac{1}{\sqrt{1 - (3^{2} t^{2})}} (3 \cdot 1)$

Step 3.6

Raise $3$ to the power of $2$ .

$6 \sin (3 t) \cos (3 t) + \frac{1}{\sqrt{1 - (9 t^{2})}} (3 \cdot 1)$

Step 3.7

Multiply $9$ by $- 1$ .

$6 \sin (3 t) \cos (3 t) + \frac{1}{\sqrt{1 - 9 t^{2}}} (3 \cdot 1)$

Step 3.8

Multiply $3$ by $1$ .

$6 \sin (3 t) \cos (3 t) + \frac{1}{\sqrt{1 - 9 t^{2}}} \cdot 3$

Step 3.9

Combine $\frac{1}{\sqrt{1 - 9 t^{2}}}$ and $3$ .

$6 \sin (3 t) \cos (3 t) + \frac{3}{\sqrt{1 - 9 t^{2}}}$

Step 4

Reorder terms.

$6 \cos (3 t) \sin (3 t) + \frac{3}{\sqrt{1 - 9 t^{2}}}$