Calculus Examples

$A e^{k t} \sin (b t)$

Step 1

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

$A \frac{d}{d t} [e^{k t} \sin (b t)]$

Step 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 $f (t) = e^{k t}$ and $g (t) = \sin (b t)$ .

$A (e^{k t} \frac{d}{d t} [\sin (b t)] + \sin (b t) \frac{d}{d t} [e^{k t}])$

Step 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 $g (t) = b t$ .

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Step 3.1

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

$A (e^{k t} (\frac{d}{d u_{1}} [\sin (u_{1})] \frac{d}{d t} [b t]) + \sin (b t) \frac{d}{d t} [e^{k t}])$

Step 3.2

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

$A (e^{k t} (\cos (u_{1}) \frac{d}{d t} [b t]) + \sin (b t) \frac{d}{d t} [e^{k t}])$

Step 3.3

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

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

Step 4

Differentiate.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Multiply $b$ by $1$ .

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

Step 5

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 $g (t) = k t$ .

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

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

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

Step 6

Differentiate.

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Step 6.1

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

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

Step 6.2

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

$A (e^{k t} (\cos (b t) b) + \sin (b t) (e^{k t} (k \cdot 1)))$

Step 6.3

Multiply $k$ by $1$ .

$A (e^{k t} (\cos (b t) b) + \sin (b t) (e^{k t} k))$

Step 7

Simplify.

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Step 7.1

Apply the distributive property.

$A (e^{k t} (\cos (b t) b)) + A (\sin (b t) (e^{k t} k))$

Step 7.2

Remove parentheses.

$A e^{k t} (\cos (b t) b) + A \sin (b t) (e^{k t} k)$

Step 7.3

Reorder terms.

$e^{k t} A b \cos (b t) + e^{k t} A k \sin (b t)$

Step 7.4

Reorder factors in $e^{k t} A b \cos (b t) + e^{k t} A k \sin (b t)$ .

$A b e^{k t} \cos (b t) + A k e^{k t} \sin (b t)$