Calculus Examples

$a (1 - e^{- k t}) + b$

Step 1

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

$\frac{d}{d t} [a (1 - e^{- k t})] + \frac{d}{d t} [b]$

Step 2

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

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Step 2.1

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

$a \frac{d}{d t} [1 - e^{- k t}] + \frac{d}{d t} [b]$

Step 2.2

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

$a (\frac{d}{d t} [1] + \frac{d}{d t} [- e^{- k t}]) + \frac{d}{d t} [b]$

Step 2.3

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

$a (0 + \frac{d}{d t} [- e^{- k t}]) + \frac{d}{d t} [b]$

Step 2.4

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

$a (0 - \frac{d}{d t} [e^{- k t}]) + \frac{d}{d t} [b]$

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 $g (t) = - k t$ .

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Step 2.5.1

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

$a (0 - (\frac{d}{d u} [e^{u}] \frac{d}{d t} [- k t])) + \frac{d}{d t} [b]$

Step 2.5.2

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

$a (0 - (e^{u} \frac{d}{d t} [- k t])) + \frac{d}{d t} [b]$

Step 2.5.3

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

$a (0 - (e^{- k t} \frac{d}{d t} [- k t])) + \frac{d}{d t} [b]$

Step 2.6

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

$a (0 - (e^{- k t} (- k \frac{d}{d t} [t]))) + \frac{d}{d t} [b]$

Step 2.7

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

$a (0 - (e^{- k t} (- k \cdot 1))) + \frac{d}{d t} [b]$

Step 2.8

Multiply $- 1$ by $1$ .

$a (0 - (e^{- k t} (- k))) + \frac{d}{d t} [b]$

Step 2.9

Multiply $- 1$ by $- 1$ .

$a (0 + 1 (e^{- k t} (k))) + \frac{d}{d t} [b]$

Step 2.10

Multiply $e^{- k t}$ by $1$ .

$a (0 + e^{- k t} k) + \frac{d}{d t} [b]$

Step 2.11

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

$a e^{- k t} k + \frac{d}{d t} [b]$

Step 3

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

$a e^{- k t} k + 0$

Step 4

Simplify.

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Step 4.1

Add $a e^{- k t} k$ and $0$ .

$a e^{- k t} k$

Step 4.2

Reorder the factors of $a e^{- k t} k$ .

$e^{- k t} a k$

Step 4.3

Reorder factors in $e^{- k t} a k$ .

$a k e^{- k t}$