Calculus Examples

$r (t) = e^{5 t} i + e^{5 t} \cos (t) j + e^{5 t} \sin (t) k$

Step 1

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

$\frac{d}{d t} [e^{5 t} i] + \frac{d}{d t} [e^{5 t} \cos (t) j] + \frac{d}{d t} [e^{5 t} \sin (t) k]$

Step 2

Evaluate $\frac{d}{d t} [e^{5 t} i]$ .

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Step 2.1

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

$i \frac{d}{d t} [e^{5 t}] + \frac{d}{d t} [e^{5 t} \cos (t) j] + \frac{d}{d t} [e^{5 t} \sin (t) k]$

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 $f (t) = e^{t}$ and $g (t) = 5 t$ .

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Step 2.2.1

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

$i (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d t} [5 t]) + \frac{d}{d t} [e^{5 t} \cos (t) j] + \frac{d}{d t} [e^{5 t} \sin (t) k]$

Step 2.2.2

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

$i (e^{u_{1}} \frac{d}{d t} [5 t]) + \frac{d}{d t} [e^{5 t} \cos (t) j] + \frac{d}{d t} [e^{5 t} \sin (t) k]$

Step 2.2.3

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

$i (e^{5 t} \frac{d}{d t} [5 t]) + \frac{d}{d t} [e^{5 t} \cos (t) j] + \frac{d}{d t} [e^{5 t} \sin (t) k]$

Step 2.3

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

$i (e^{5 t} (5 \frac{d}{d t} [t])) + \frac{d}{d t} [e^{5 t} \cos (t) j] + \frac{d}{d t} [e^{5 t} \sin (t) k]$

Step 2.4

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

$i (e^{5 t} (5 \cdot 1)) + \frac{d}{d t} [e^{5 t} \cos (t) j] + \frac{d}{d t} [e^{5 t} \sin (t) k]$

Step 2.5

Multiply $5$ by $1$ .

$i (e^{5 t} \cdot 5) + \frac{d}{d t} [e^{5 t} \cos (t) j] + \frac{d}{d t} [e^{5 t} \sin (t) k]$

Step 2.6

Move $5$ to the left of $e^{5 t}$ .

$i (5 \cdot e^{5 t}) + \frac{d}{d t} [e^{5 t} \cos (t) j] + \frac{d}{d t} [e^{5 t} \sin (t) k]$

Step 2.7

Move $5$ to the left of $i$ .

$5 i e^{5 t} + \frac{d}{d t} [e^{5 t} \cos (t) j] + \frac{d}{d t} [e^{5 t} \sin (t) k]$

Step 3

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

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Step 3.1

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

$5 i e^{5 t} + j \frac{d}{d t} [e^{5 t} \cos (t)] + \frac{d}{d t} [e^{5 t} \sin (t) k]$

Step 3.2

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 $f (t) = e^{5 t}$ and $g (t) = \cos (t)$ .

$5 i e^{5 t} + j (e^{5 t} \frac{d}{d t} [\cos (t)] + \cos (t) \frac{d}{d t} [e^{5 t}]) + \frac{d}{d t} [e^{5 t} \sin (t) k]$

Step 3.3

The derivative of $\cos (t)$ with respect to $t$ is $- \sin (t)$ .

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

Step 3.4

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 $g (t) = 5 t$ .

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Step 3.4.1

To apply the Chain Rule, set $u_{2}$ as $5 t$ .

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

Step 3.4.2

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

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

Step 3.4.3

Replace all occurrences of $u_{2}$ with $5 t$ .

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

Step 3.5

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

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

Step 3.6

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

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

Step 3.7

Multiply $5$ by $1$ .

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

Step 3.8

Move $5$ to the left of $e^{5 t}$ .

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

Step 4

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

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

The derivative of $\sin (t)$ with respect to $t$ is $\cos (t)$ .

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

Step 4.4

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 $g (t) = 5 t$ .

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Step 4.4.1

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

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

Step 4.4.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{3}} [a^{u_{3}}]$ is $a^{u_{3}} \ln (a)$ where $a$ = $e$ .

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

Step 4.4.3

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

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

Step 4.5

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

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

Step 4.6

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

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

Step 4.7

Multiply $5$ by $1$ .

$5 i e^{5 t} + j (e^{5 t} (- \sin (t)) + \cos (t) (5 e^{5 t})) + k (e^{5 t} \cos (t) + \sin (t) (e^{5 t} \cdot 5))$

Step 4.8

Move $5$ to the left of $e^{5 t}$ .

$5 i e^{5 t} + j (e^{5 t} (- \sin (t)) + \cos (t) (5 e^{5 t})) + k (e^{5 t} \cos (t) + \sin (t) (5 e^{5 t}))$

Step 5

Simplify.

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Step 5.1

Apply the distributive property.

$5 i e^{5 t} + j (e^{5 t} (- \sin (t))) + j (\cos (t) (5 e^{5 t})) + k (e^{5 t} \cos (t) + \sin (t) (5 e^{5 t}))$

Step 5.2

Apply the distributive property.

$5 i e^{5 t} + j (e^{5 t} (- \sin (t))) + j (\cos (t) (5 e^{5 t})) + k (e^{5 t} \cos (t)) + k (\sin (t) (5 e^{5 t}))$

Step 5.3

Remove unnecessary parentheses.

$5 i e^{5 t} + j e^{5 t} (- \sin (t)) + j \cos (t) (5 e^{5 t}) + k e^{5 t} \cos (t) + k \sin (t) (5 e^{5 t})$

Step 5.4

Reorder terms.

$5 i e^{5 t} - e^{5 t} j \sin (t) + 5 e^{5 t} j \cos (t) + e^{5 t} k \cos (t) + 5 e^{5 t} k \sin (t)$

Step 5.5

Reorder factors in $5 i e^{5 t} - e^{5 t} j \sin (t) + 5 e^{5 t} j \cos (t) + e^{5 t} k \cos (t) + 5 e^{5 t} k \sin (t)$ .

$5 i e^{5 t} - j e^{5 t} \sin (t) + 5 j e^{5 t} \cos (t) + k e^{5 t} \cos (t) + 5 k e^{5 t} \sin (t)$