Calculus Examples

$- 2 e^{3 x} \sin (2 x) + 3 e^{3 x} \cos (2 x)$

Step 1

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

$\frac{d}{d x} [- 2 e^{3 x} \sin (2 x)] + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2

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

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Step 2.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 e^{3 x} \sin (2 x)$ with respect to $x$ is $- 2 \frac{d}{d x} [e^{3 x} \sin (2 x)]$ .

$- 2 \frac{d}{d x} [e^{3 x} \sin (2 x)] + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = e^{3 x}$ and $g (x) = \sin (2 x)$ .

$- 2 (e^{3 x} \frac{d}{d x} [\sin (2 x)] + \sin (2 x) \frac{d}{d x} [e^{3 x}]) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = 2 x$ .

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Step 2.3.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

$- 2 (e^{3 x} (\frac{d}{d u_{1}} [\sin (u_{1})] \frac{d}{d x} [2 x]) + \sin (2 x) \frac{d}{d x} [e^{3 x}]) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.3.2

The derivative of $\sin (u_{1})$ with respect to $u_{1}$ is $\cos (u_{1})$ .

$- 2 (e^{3 x} (\cos (u_{1}) \frac{d}{d x} [2 x]) + \sin (2 x) \frac{d}{d x} [e^{3 x}]) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.3.3

Replace all occurrences of $u_{1}$ with $2 x$ .

$- 2 (e^{3 x} (\cos (2 x) \frac{d}{d x} [2 x]) + \sin (2 x) \frac{d}{d x} [e^{3 x}]) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.4

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

$- 2 (e^{3 x} (\cos (2 x) (2 \frac{d}{d x} [x])) + \sin (2 x) \frac{d}{d x} [e^{3 x}]) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.5

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

$- 2 (e^{3 x} (\cos (2 x) (2 \cdot 1)) + \sin (2 x) \frac{d}{d x} [e^{3 x}]) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.6

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 3 x$ .

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Step 2.6.1

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

$- 2 (e^{3 x} (\cos (2 x) (2 \cdot 1)) + \sin (2 x) (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [3 x])) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.6.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$- 2 (e^{3 x} (\cos (2 x) (2 \cdot 1)) + \sin (2 x) (e^{u_{2}} \frac{d}{d x} [3 x])) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.6.3

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

$- 2 (e^{3 x} (\cos (2 x) (2 \cdot 1)) + \sin (2 x) (e^{3 x} \frac{d}{d x} [3 x])) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.7

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

$- 2 (e^{3 x} (\cos (2 x) (2 \cdot 1)) + \sin (2 x) (e^{3 x} (3 \frac{d}{d x} [x]))) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.8

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

$- 2 (e^{3 x} (\cos (2 x) (2 \cdot 1)) + \sin (2 x) (e^{3 x} (3 \cdot 1))) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.9

Multiply $2$ by $1$ .

$- 2 (e^{3 x} (\cos (2 x) \cdot 2) + \sin (2 x) (e^{3 x} (3 \cdot 1))) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.10

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

$- 2 (e^{3 x} (2 \cdot \cos (2 x)) + \sin (2 x) (e^{3 x} (3 \cdot 1))) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.11

Multiply $3$ by $1$ .

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (e^{3 x} \cdot 3)) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 2.12

Move $3$ to the left of $e^{3 x}$ .

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + \frac{d}{d x} [3 e^{3 x} \cos (2 x)]$

Step 3

Evaluate $\frac{d}{d x} [3 e^{3 x} \cos (2 x)]$ .

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Step 3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 e^{3 x} \cos (2 x)$ with respect to $x$ is $3 \frac{d}{d x} [e^{3 x} \cos (2 x)]$ .

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 \frac{d}{d x} [e^{3 x} \cos (2 x)]$

Step 3.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = e^{3 x}$ and $g (x) = \cos (2 x)$ .

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} \frac{d}{d x} [\cos (2 x)] + \cos (2 x) \frac{d}{d x} [e^{3 x}])$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = 2 x$ .

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Step 3.3.1

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

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (\frac{d}{d u_{3}} [\cos (u_{3})] \frac{d}{d x} [2 x]) + \cos (2 x) \frac{d}{d x} [e^{3 x}])$

Step 3.3.2

The derivative of $\cos (u_{3})$ with respect to $u_{3}$ is $- \sin (u_{3})$ .

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- \sin (u_{3}) \frac{d}{d x} [2 x]) + \cos (2 x) \frac{d}{d x} [e^{3 x}])$

Step 3.3.3

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

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- \sin (2 x) \frac{d}{d x} [2 x]) + \cos (2 x) \frac{d}{d x} [e^{3 x}])$

Step 3.4

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

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- \sin (2 x) (2 \frac{d}{d x} [x])) + \cos (2 x) \frac{d}{d x} [e^{3 x}])$

Step 3.5

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

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- \sin (2 x) (2 \cdot 1)) + \cos (2 x) \frac{d}{d x} [e^{3 x}])$

Step 3.6

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 3 x$ .

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Step 3.6.1

To apply the Chain Rule, set $u_{4}$ as $3 x$ .

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- \sin (2 x) (2 \cdot 1)) + \cos (2 x) (\frac{d}{d u_{4}} [e^{u_{4}}] \frac{d}{d x} [3 x]))$

Step 3.6.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{4}} [a^{u_{4}}]$ is $a^{u_{4}} \ln (a)$ where $a$ = $e$ .

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- \sin (2 x) (2 \cdot 1)) + \cos (2 x) (e^{u_{4}} \frac{d}{d x} [3 x]))$

Step 3.6.3

Replace all occurrences of $u_{4}$ with $3 x$ .

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- \sin (2 x) (2 \cdot 1)) + \cos (2 x) (e^{3 x} \frac{d}{d x} [3 x]))$

Step 3.7

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

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- \sin (2 x) (2 \cdot 1)) + \cos (2 x) (e^{3 x} (3 \frac{d}{d x} [x])))$

Step 3.8

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

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- \sin (2 x) (2 \cdot 1)) + \cos (2 x) (e^{3 x} (3 \cdot 1)))$

Step 3.9

Multiply $2$ by $1$ .

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- \sin (2 x) \cdot 2) + \cos (2 x) (e^{3 x} (3 \cdot 1)))$

Step 3.10

Multiply $2$ by $- 1$ .

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- 2 \sin (2 x)) + \cos (2 x) (e^{3 x} (3 \cdot 1)))$

Step 3.11

Multiply $3$ by $1$ .

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- 2 \sin (2 x)) + \cos (2 x) (e^{3 x} \cdot 3))$

Step 3.12

Move $3$ to the left of $e^{3 x}$ .

$- 2 (e^{3 x} (2 \cos (2 x)) + \sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- 2 \sin (2 x)) + \cos (2 x) (3 e^{3 x}))$

Step 4

Simplify.

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Step 4.1

Apply the distributive property.

$- 2 (e^{3 x} (2 \cos (2 x))) - 2 (\sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- 2 \sin (2 x)) + \cos (2 x) (3 e^{3 x}))$

Step 4.2

Apply the distributive property.

$- 2 (e^{3 x} (2 \cos (2 x))) - 2 (\sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- 2 \sin (2 x))) + 3 (\cos (2 x) (3 e^{3 x}))$

Step 4.3

Combine terms.

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Step 4.3.1

Multiply $2$ by $- 2$ .

$- 4 (e^{3 x} (\cos (2 x))) - 2 (\sin (2 x) (3 e^{3 x})) + 3 (e^{3 x} (- 2 \sin (2 x))) + 3 (\cos (2 x) (3 e^{3 x}))$

Step 4.3.2

Multiply $3$ by $- 2$ .

$- 4 e^{3 x} \cos (2 x) - 6 (\sin (2 x) (e^{3 x})) + 3 (e^{3 x} (- 2 \sin (2 x))) + 3 (\cos (2 x) (3 e^{3 x}))$

Step 4.3.3

Multiply $- 2$ by $3$ .

$- 4 e^{3 x} \cos (2 x) - 6 \sin (2 x) e^{3 x} - 6 (e^{3 x} (\sin (2 x))) + 3 (\cos (2 x) (3 e^{3 x}))$

Step 4.3.4

Multiply $3$ by $3$ .

$- 4 e^{3 x} \cos (2 x) - 6 \sin (2 x) e^{3 x} - 6 e^{3 x} \sin (2 x) + 9 (\cos (2 x) (e^{3 x}))$

Step 4.3.5

Subtract $6 e^{3 x} \sin (2 x)$ from $- 6 \sin (2 x) e^{3 x}$ .

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Step 4.3.5.1

Move $\sin (2 x)$ .

$- 4 e^{3 x} \cos (2 x) - 6 e^{3 x} \sin (2 x) - 6 e^{3 x} \sin (2 x) + 9 \cos (2 x) e^{3 x}$

Step 4.3.5.2

Subtract $6 e^{3 x} \sin (2 x)$ from $- 6 e^{3 x} \sin (2 x)$ .

$- 4 e^{3 x} \cos (2 x) - 12 e^{3 x} \sin (2 x) + 9 \cos (2 x) e^{3 x}$

Step 4.3.6

Add $- 4 e^{3 x} \cos (2 x)$ and $9 \cos (2 x) e^{3 x}$ .

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Step 4.3.6.1

Move $\cos (2 x)$ .

$- 4 e^{3 x} \cos (2 x) + 9 e^{3 x} \cos (2 x) - 12 e^{3 x} \sin (2 x)$

Step 4.3.6.2

Add $- 4 e^{3 x} \cos (2 x)$ and $9 e^{3 x} \cos (2 x)$ .

$5 e^{3 x} \cos (2 x) - 12 e^{3 x} \sin (2 x)$