Calculus Examples

$f (t) = e^{2 t \sin (2 t)}$

Step 1

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

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $2 t \sin (2 t)$ .

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

Step 1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 1.3

Replace all occurrences of $u_{1}$ with $2 t \sin (2 t)$ .

$e^{2 t \sin (2 t)} \frac{d}{d t} [2 t \sin (2 t)]$

Step 2

Differentiate using the Constant Multiple Rule.

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Step 2.1

Since $2$ is constant with respect to $t$ , the derivative of $2 t \sin (2 t)$ with respect to $t$ is $2 \frac{d}{d t} [t \sin (2 t)]$ .

$e^{2 t \sin (2 t)} (2 \frac{d}{d t} [t \sin (2 t)])$

Step 2.2

Move $2$ to the left of $e^{2 t \sin (2 t)}$ .

$2 \cdot e^{2 t \sin (2 t)} \frac{d}{d t} [t \sin (2 t)]$

Step 3

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 $g (t) = \sin (2 t)$ .

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

Step 4

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 $g (t) = 2 t$ .

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

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

$2 e^{2 t \sin (2 t)} (t (\cos (2 t) \frac{d}{d t} [2 t]) + \sin (2 t) \frac{d}{d t} [t])$

Step 5

Differentiate.

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Step 5.1

Since $2$ is constant with respect to $t$ , the derivative of $2 t$ with respect to $t$ is $2 \frac{d}{d t} [t]$ .

$2 e^{2 t \sin (2 t)} (t (\cos (2 t) (2 \frac{d}{d t} [t])) + \sin (2 t) \frac{d}{d t} [t])$

Step 5.2

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

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

Step 5.3

Simplify the expression.

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Step 5.3.1

Multiply $2$ by $1$ .

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

Step 5.3.2

Move $2$ to the left of $\cos (2 t)$ .

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

Step 5.4

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

$2 e^{2 t \sin (2 t)} (t (2 \cos (2 t)) + \sin (2 t) \cdot 1)$

Step 5.5

Multiply $\sin (2 t)$ by $1$ .

$2 e^{2 t \sin (2 t)} (t (2 \cos (2 t)) + \sin (2 t))$

Step 6

Simplify.

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Step 6.1

Apply the distributive property.

$2 e^{2 t \sin (2 t)} (t (2 \cos (2 t))) + 2 e^{2 t \sin (2 t)} \sin (2 t)$

Step 6.2

Multiply $2$ by $2$ .

$4 e^{2 t \sin (2 t)} t \cos (2 t) + 2 e^{2 t \sin (2 t)} \sin (2 t)$