Calculus Examples

$h (t) = e^{- 0.3 t} \sin (5 t)$

Step 1

Find the first derivative.

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Step 1.1

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 $f (t) = e^{- 0.3 t}$ and $g (t) = \sin (5 t)$ .

$e^{- 0.3 t} \frac{d}{d t} [\sin (5 t)] + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}]$

Step 1.2

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 $g (t) = 5 t$ .

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Step 1.2.1

To apply the Chain Rule, set $u_{1}$ as $5 t$ .

$e^{- 0.3 t} (\frac{d}{d u_{1}} [\sin (u_{1})] \frac{d}{d t} [5 t]) + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}]$

Step 1.2.2

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

$e^{- 0.3 t} (\cos (u_{1}) \frac{d}{d t} [5 t]) + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}]$

Step 1.2.3

Replace all occurrences of $u_{1}$ with $5 t$ .

$e^{- 0.3 t} (\cos (5 t) \frac{d}{d t} [5 t]) + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}]$

Step 1.3

Differentiate.

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Step 1.3.1

Since $5$ is constant with respect to $t$ , the derivative of $5 t$ with respect to $t$ is $5 \frac{d}{d t} [t]$ .

$e^{- 0.3 t} (\cos (5 t) (5 \frac{d}{d t} [t])) + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}]$

Step 1.3.2

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

$e^{- 0.3 t} (\cos (5 t) (5 \cdot 1)) + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}]$

Step 1.3.3

Simplify the expression.

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Step 1.3.3.1

Multiply $5$ by $1$ .

$e^{- 0.3 t} (\cos (5 t) \cdot 5) + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}]$

Step 1.3.3.2

Move $5$ to the left of $\cos (5 t)$ .

$e^{- 0.3 t} (5 \cdot \cos (5 t)) + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}]$

Step 1.4

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - 0.3 t$ .

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Step 1.4.1

To apply the Chain Rule, set $u_{2}$ as $- 0.3 t$ .

$e^{- 0.3 t} (5 \cos (5 t)) + \sin (5 t) (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d t} [- 0.3 t])$

Step 1.4.2

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

$e^{- 0.3 t} (5 \cos (5 t)) + \sin (5 t) (e^{u_{2}} \frac{d}{d t} [- 0.3 t])$

Step 1.4.3

Replace all occurrences of $u_{2}$ with $- 0.3 t$ .

$e^{- 0.3 t} (5 \cos (5 t)) + \sin (5 t) (e^{- 0.3 t} \frac{d}{d t} [- 0.3 t])$

Step 1.5

Differentiate.

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Step 1.5.1

Since $- 0.3$ is constant with respect to $t$ , the derivative of $- 0.3 t$ with respect to $t$ is $- 0.3 \frac{d}{d t} [t]$ .

$e^{- 0.3 t} (5 \cos (5 t)) + \sin (5 t) (e^{- 0.3 t} (- 0.3 \frac{d}{d t} [t]))$

Step 1.5.2

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

$e^{- 0.3 t} (5 \cos (5 t)) + \sin (5 t) (e^{- 0.3 t} (- 0.3 \cdot 1))$

Step 1.5.3

Simplify the expression.

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Step 1.5.3.1

Multiply $- 0.3$ by $1$ .

$e^{- 0.3 t} (5 \cos (5 t)) + \sin (5 t) (e^{- 0.3 t} \cdot - 0.3)$

Step 1.5.3.2

Move $- 0.3$ to the left of $e^{- 0.3 t}$ .

$e^{- 0.3 t} (5 \cos (5 t)) + \sin (5 t) (- 0.3 \cdot e^{- 0.3 t})$

Step 1.5.3.3

Reorder terms.

$f' (t) = 5 e^{- 0.3 t} \cos (5 t) - 0.3 e^{- 0.3 t} \sin (5 t)$

Step 2

Find the second derivative.

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Step 2.1

By the Sum Rule, the derivative of $5 e^{- 0.3 t} \cos (5 t) - 0.3 e^{- 0.3 t} \sin (5 t)$ with respect to $t$ is $\frac{d}{d t} [5 e^{- 0.3 t} \cos (5 t)] + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$ .

$\frac{d}{d t} [5 e^{- 0.3 t} \cos (5 t)] + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2

Evaluate $\frac{d}{d t} [5 e^{- 0.3 t} \cos (5 t)]$ .

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Step 2.2.1

Since $5$ is constant with respect to $t$ , the derivative of $5 e^{- 0.3 t} \cos (5 t)$ with respect to $t$ is $5 \frac{d}{d t} [e^{- 0.3 t} \cos (5 t)]$ .

$5 \frac{d}{d t} [e^{- 0.3 t} \cos (5 t)] + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.2

$5 (e^{- 0.3 t} \frac{d}{d t} [\cos (5 t)] + \cos (5 t) \frac{d}{d t} [e^{- 0.3 t}]) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \cos (t)$ and $g (t) = 5 t$ .

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Step 2.2.3.1

To apply the Chain Rule, set $u_{1}$ as $5 t$ .

$5 (e^{- 0.3 t} (\frac{d}{d u_{1}} [\cos (u_{1})] \frac{d}{d t} [5 t]) + \cos (5 t) \frac{d}{d t} [e^{- 0.3 t}]) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.3.2

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

$5 (e^{- 0.3 t} (- \sin (u_{1}) \frac{d}{d t} [5 t]) + \cos (5 t) \frac{d}{d t} [e^{- 0.3 t}]) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.3.3

Replace all occurrences of $u_{1}$ with $5 t$ .

$5 (e^{- 0.3 t} (- \sin (5 t) \frac{d}{d t} [5 t]) + \cos (5 t) \frac{d}{d t} [e^{- 0.3 t}]) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.4

Since $5$ is constant with respect to $t$ , the derivative of $5 t$ with respect to $t$ is $5 \frac{d}{d t} [t]$ .

$5 (e^{- 0.3 t} (- \sin (5 t) (5 \frac{d}{d t} [t])) + \cos (5 t) \frac{d}{d t} [e^{- 0.3 t}]) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.5

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

$5 (e^{- 0.3 t} (- \sin (5 t) (5 \cdot 1)) + \cos (5 t) \frac{d}{d t} [e^{- 0.3 t}]) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.6

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 $g (t) = - 0.3 t$ .

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Step 2.2.6.1

To apply the Chain Rule, set $u_{2}$ as $- 0.3 t$ .

$5 (e^{- 0.3 t} (- \sin (5 t) (5 \cdot 1)) + \cos (5 t) (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d t} [- 0.3 t])) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.6.2

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

$5 (e^{- 0.3 t} (- \sin (5 t) (5 \cdot 1)) + \cos (5 t) (e^{u_{2}} \frac{d}{d t} [- 0.3 t])) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.6.3

Replace all occurrences of $u_{2}$ with $- 0.3 t$ .

$5 (e^{- 0.3 t} (- \sin (5 t) (5 \cdot 1)) + \cos (5 t) (e^{- 0.3 t} \frac{d}{d t} [- 0.3 t])) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.7

Since $- 0.3$ is constant with respect to $t$ , the derivative of $- 0.3 t$ with respect to $t$ is $- 0.3 \frac{d}{d t} [t]$ .

$5 (e^{- 0.3 t} (- \sin (5 t) (5 \cdot 1)) + \cos (5 t) (e^{- 0.3 t} (- 0.3 \frac{d}{d t} [t]))) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.8

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

$5 (e^{- 0.3 t} (- \sin (5 t) (5 \cdot 1)) + \cos (5 t) (e^{- 0.3 t} (- 0.3 \cdot 1))) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.9

Multiply $5$ by $1$ .

$5 (e^{- 0.3 t} (- \sin (5 t) \cdot 5) + \cos (5 t) (e^{- 0.3 t} (- 0.3 \cdot 1))) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.10

Multiply $5$ by $- 1$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (e^{- 0.3 t} (- 0.3 \cdot 1))) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.11

Multiply $- 0.3$ by $1$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (e^{- 0.3 t} \cdot - 0.3)) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.2.12

Move $- 0.3$ to the left of $e^{- 0.3 t}$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) + \frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$

Step 2.3

Evaluate $\frac{d}{d t} [- 0.3 e^{- 0.3 t} \sin (5 t)]$ .

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Step 2.3.1

Since $- 0.3$ is constant with respect to $t$ , the derivative of $- 0.3 e^{- 0.3 t} \sin (5 t)$ with respect to $t$ is $- 0.3 \frac{d}{d t} [e^{- 0.3 t} \sin (5 t)]$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 \frac{d}{d t} [e^{- 0.3 t} \sin (5 t)]$

Step 2.3.2

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} \frac{d}{d t} [\sin (5 t)] + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}])$

Step 2.3.3

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 $g (t) = 5 t$ .

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Step 2.3.3.1

To apply the Chain Rule, set $u_{3}$ as $5 t$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (\frac{d}{d u_{3}} [\sin (u_{3})] \frac{d}{d t} [5 t]) + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}])$

Step 2.3.3.2

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

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (\cos (u_{3}) \frac{d}{d t} [5 t]) + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}])$

Step 2.3.3.3

Replace all occurrences of $u_{3}$ with $5 t$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (\cos (5 t) \frac{d}{d t} [5 t]) + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}])$

Step 2.3.4

Since $5$ is constant with respect to $t$ , the derivative of $5 t$ with respect to $t$ is $5 \frac{d}{d t} [t]$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (\cos (5 t) (5 \frac{d}{d t} [t])) + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}])$

Step 2.3.5

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

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (\cos (5 t) (5 \cdot 1)) + \sin (5 t) \frac{d}{d t} [e^{- 0.3 t}])$

Step 2.3.6

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 $g (t) = - 0.3 t$ .

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Step 2.3.6.1

To apply the Chain Rule, set $u_{4}$ as $- 0.3 t$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (\cos (5 t) (5 \cdot 1)) + \sin (5 t) (\frac{d}{d u_{4}} [e^{u_{4}}] \frac{d}{d t} [- 0.3 t]))$

Step 2.3.6.2

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

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (\cos (5 t) (5 \cdot 1)) + \sin (5 t) (e^{u_{4}} \frac{d}{d t} [- 0.3 t]))$

Step 2.3.6.3

Replace all occurrences of $u_{4}$ with $- 0.3 t$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (\cos (5 t) (5 \cdot 1)) + \sin (5 t) (e^{- 0.3 t} \frac{d}{d t} [- 0.3 t]))$

Step 2.3.7

Since $- 0.3$ is constant with respect to $t$ , the derivative of $- 0.3 t$ with respect to $t$ is $- 0.3 \frac{d}{d t} [t]$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (\cos (5 t) (5 \cdot 1)) + \sin (5 t) (e^{- 0.3 t} (- 0.3 \frac{d}{d t} [t])))$

Step 2.3.8

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

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (\cos (5 t) (5 \cdot 1)) + \sin (5 t) (e^{- 0.3 t} (- 0.3 \cdot 1)))$

Step 2.3.9

Multiply $5$ by $1$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (\cos (5 t) \cdot 5) + \sin (5 t) (e^{- 0.3 t} (- 0.3 \cdot 1)))$

Step 2.3.10

Move $5$ to the left of $\cos (5 t)$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (5 \cdot \cos (5 t)) + \sin (5 t) (e^{- 0.3 t} (- 0.3 \cdot 1)))$

Step 2.3.11

Multiply $- 0.3$ by $1$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (5 \cos (5 t)) + \sin (5 t) (e^{- 0.3 t} \cdot - 0.3))$

Step 2.3.12

Move $- 0.3$ to the left of $e^{- 0.3 t}$ .

$5 (e^{- 0.3 t} (- 5 \sin (5 t)) + \cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (5 \cos (5 t)) + \sin (5 t) (- 0.3 e^{- 0.3 t}))$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$5 (e^{- 0.3 t} (- 5 \sin (5 t))) + 5 (\cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (5 \cos (5 t)) + \sin (5 t) (- 0.3 e^{- 0.3 t}))$

Step 2.4.2

Apply the distributive property.

$5 (e^{- 0.3 t} (- 5 \sin (5 t))) + 5 (\cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (5 \cos (5 t))) - 0.3 (\sin (5 t) (- 0.3 e^{- 0.3 t}))$

Step 2.4.3

Combine terms.

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Step 2.4.3.1

Multiply $- 5$ by $5$ .

$- 25 (e^{- 0.3 t} (\sin (5 t))) + 5 (\cos (5 t) (- 0.3 e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (5 \cos (5 t))) - 0.3 (\sin (5 t) (- 0.3 e^{- 0.3 t}))$

Step 2.4.3.2

Multiply $- 0.3$ by $5$ .

$- 25 e^{- 0.3 t} \sin (5 t) - 1.5 (\cos (5 t) (e^{- 0.3 t})) - 0.3 (e^{- 0.3 t} (5 \cos (5 t))) - 0.3 (\sin (5 t) (- 0.3 e^{- 0.3 t}))$

Step 2.4.3.3

Multiply $5$ by $- 0.3$ .

$- 25 e^{- 0.3 t} \sin (5 t) - 1.5 \cos (5 t) e^{- 0.3 t} - 1.5 (e^{- 0.3 t} (\cos (5 t))) - 0.3 (\sin (5 t) (- 0.3 e^{- 0.3 t}))$

Step 2.4.3.4

Multiply $- 0.3$ by $- 0.3$ .

$- 25 e^{- 0.3 t} \sin (5 t) - 1.5 \cos (5 t) e^{- 0.3 t} - 1.5 e^{- 0.3 t} \cos (5 t) + 0.09 (\sin (5 t) (e^{- 0.3 t}))$

Step 2.4.3.5

Subtract $1.5 e^{- 0.3 t} \cos (5 t)$ from $- 1.5 \cos (5 t) e^{- 0.3 t}$ .

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Step 2.4.3.5.1

Move $\cos (5 t)$ .

$- 25 e^{- 0.3 t} \sin (5 t) - 1.5 e^{- 0.3 t} \cos (5 t) - 1.5 e^{- 0.3 t} \cos (5 t) + 0.09 \sin (5 t) e^{- 0.3 t}$

Step 2.4.3.5.2

Subtract $1.5 e^{- 0.3 t} \cos (5 t)$ from $- 1.5 e^{- 0.3 t} \cos (5 t)$ .

$- 25 e^{- 0.3 t} \sin (5 t) - 3 e^{- 0.3 t} \cos (5 t) + 0.09 \sin (5 t) e^{- 0.3 t}$

Step 2.4.3.6

Add $- 25 e^{- 0.3 t} \sin (5 t)$ and $0.09 \sin (5 t) e^{- 0.3 t}$ .

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Step 2.4.3.6.1

Move $\sin (5 t)$ .

$- 25 e^{- 0.3 t} \sin (5 t) + 0.09 e^{- 0.3 t} \sin (5 t) - 3 e^{- 0.3 t} \cos (5 t)$

Step 2.4.3.6.2

Add $- 25 e^{- 0.3 t} \sin (5 t)$ and $0.09 e^{- 0.3 t} \sin (5 t)$ .

$f'' (t) = - 24.91 e^{- 0.3 t} \sin (5 t) - 3 e^{- 0.3 t} \cos (5 t)$

Step 3

The second derivative of $h (t)$ with respect to $t$ is $- 24.91 e^{- 0.3 t} \sin (5 t) - 3 e^{- 0.3 t} \cos (5 t)$ .

$- 24.91 e^{- 0.3 t} \sin (5 t) - 3 e^{- 0.3 t} \cos (5 t)$