Calculus Examples

$V (t) = 45 (1 - e^{- 1.6 t}) + 31$

Step 1

By the Sum Rule, the derivative of $45 (1 - e^{- 1.6 t}) + 31$ with respect to $t$ is $\frac{d}{d t} [45 (1 - e^{- 1.6 t})] + \frac{d}{d t} [31]$ .

$\frac{d}{d t} [45 (1 - e^{- 1.6 t})] + \frac{d}{d t} [31]$

Step 2

Evaluate $\frac{d}{d t} [45 (1 - e^{- 1.6 t})]$ .

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Step 2.1

Since $45$ is constant with respect to $t$ , the derivative of $45 (1 - e^{- 1.6 t})$ with respect to $t$ is $45 \frac{d}{d t} [1 - e^{- 1.6 t}]$ .

$45 \frac{d}{d t} [1 - e^{- 1.6 t}] + \frac{d}{d t} [31]$

Step 2.2

By the Sum Rule, the derivative of $1 - e^{- 1.6 t}$ with respect to $t$ is $\frac{d}{d t} [1] + \frac{d}{d t} [- e^{- 1.6 t}]$ .

$45 (\frac{d}{d t} [1] + \frac{d}{d t} [- e^{- 1.6 t}]) + \frac{d}{d t} [31]$

Step 2.3

Since $1$ is constant with respect to $t$ , the derivative of $1$ with respect to $t$ is $0$ .

$45 (0 + \frac{d}{d t} [- e^{- 1.6 t}]) + \frac{d}{d t} [31]$

Step 2.4

Since $- 1$ is constant with respect to $t$ , the derivative of $- e^{- 1.6 t}$ with respect to $t$ is $- \frac{d}{d t} [e^{- 1.6 t}]$ .

$45 (0 - \frac{d}{d t} [e^{- 1.6 t}]) + \frac{d}{d t} [31]$

Step 2.5

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 $g (t) = - 1.6 t$ .

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Step 2.5.1

To apply the Chain Rule, set $u$ as $- 1.6 t$ .

$45 (0 - (\frac{d}{d u} [e^{u}] \frac{d}{d t} [- 1.6 t])) + \frac{d}{d t} [31]$

Step 2.5.2

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

$45 (0 - (e^{u} \frac{d}{d t} [- 1.6 t])) + \frac{d}{d t} [31]$

Step 2.5.3

Replace all occurrences of $u$ with $- 1.6 t$ .

$45 (0 - (e^{- 1.6 t} \frac{d}{d t} [- 1.6 t])) + \frac{d}{d t} [31]$

Step 2.6

Since $- 1.6$ is constant with respect to $t$ , the derivative of $- 1.6 t$ with respect to $t$ is $- 1.6 \frac{d}{d t} [t]$ .

$45 (0 - (e^{- 1.6 t} (- 1.6 \frac{d}{d t} [t]))) + \frac{d}{d t} [31]$

Step 2.7

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

$45 (0 - (e^{- 1.6 t} (- 1.6 \cdot 1))) + \frac{d}{d t} [31]$

Step 2.8

Multiply $- 1.6$ by $1$ .

$45 (0 - (e^{- 1.6 t} \cdot - 1.6)) + \frac{d}{d t} [31]$

Step 2.9

Move $- 1.6$ to the left of $e^{- 1.6 t}$ .

$45 (0 - (- 1.6 \cdot e^{- 1.6 t})) + \frac{d}{d t} [31]$

Step 2.10

Multiply $- 1.6$ by $- 1$ .

$45 (0 + 1.6 e^{- 1.6 t}) + \frac{d}{d t} [31]$

Step 2.11

Add $0$ and $1.6 e^{- 1.6 t}$ .

$45 (1.6 e^{- 1.6 t}) + \frac{d}{d t} [31]$

Step 2.12

Multiply $1.6$ by $45$ .

$72 e^{- 1.6 t} + \frac{d}{d t} [31]$

Step 3

Differentiate using the Constant Rule.

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Step 3.1

Since $31$ is constant with respect to $t$ , the derivative of $31$ with respect to $t$ is $0$ .

$72 e^{- 1.6 t} + 0$

Step 3.2

Add $72 e^{- 1.6 t}$ and $0$ .

$72 e^{- 1.6 t}$