Calculus Examples

g (t) = \frac{0.00331}{0.00331 + 0.99669 e^{- 3.8 t}}

$g (t) = \frac{0.00331}{0.00331 + 0.99669 e^{- 3.8 t}}$

Step 1

Differentiate using the Constant Multiple Rule.

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Step 1.1

Since

0.00331

$0.00331$ is constant with respect to

t

$t$ , the derivative of

\frac{0.00331}{0.00331 + 0.99669 e^{- 3.8 t}}

$\frac{0.00331}{0.00331 + 0.99669 e^{- 3.8 t}}$ with respect to

t

$t$ is

0.00331 \frac{d}{d t} [\frac{1}{0.00331 + 0.99669 e^{- 3.8 t}}]

$0.00331 \frac{d}{d t} [\frac{1}{0.00331 + 0.99669 e^{- 3.8 t}}]$ .

0.00331 \frac{d}{d t} [\frac{1}{0.00331 + 0.99669 e^{- 3.8 t}}]

$0.00331 \frac{d}{d t} [\frac{1}{0.00331 + 0.99669 e^{- 3.8 t}}]$

Step 1.2

Rewrite

\frac{1}{0.00331 + 0.99669 e^{- 3.8 t}}

$\frac{1}{0.00331 + 0.99669 e^{- 3.8 t}}$ as

{(0.00331 + 0.99669 e^{- 3.8 t})}^{- 1}

${(0.00331 + 0.99669 e^{- 3.8 t})}^{- 1}$ .

0.00331 \frac{d}{d t} [{(0.00331 + 0.99669 e^{- 3.8 t})}^{- 1}]

$0.00331 \frac{d}{d t} [{(0.00331 + 0.99669 e^{- 3.8 t})}^{- 1}]$

0.00331 \frac{d}{d t} [{(0.00331 + 0.99669 e^{- 3.8 t})}^{- 1}]

$0.00331 \frac{d}{d t} [{(0.00331 + 0.99669 e^{- 3.8 t})}^{- 1}]$

Step 2

Differentiate using the chain rule, which states that

\frac{d}{d t} [f (g (t))]

$\frac{d}{d t} [f (g (t))]$ is

$f' (g (t)) g' (t)$ where

$f (t) = t^{- 1}$ and

$g (t) = 0.00331 + 0.99669 e^{- 3.8 t}$ .

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Step 2.1

To apply the Chain Rule, set

$u_{1}$ as

$0.00331 + 0.99669 e^{- 3.8 t}$ .

$0.00331 (\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d t} [0.00331 + 0.99669 e^{- 3.8 t}])$

Step 2.2

Differentiate using the Power Rule which states that

$\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is

$n {u_{1}}^{n - 1}$ where

$n = - 1$ .

$0.00331 (- {u_{1}}^{- 2} \frac{d}{d t} [0.00331 + 0.99669 e^{- 3.8 t}])$

Step 2.3

Replace all occurrences of

$u_{1}$ with

$0.00331 + 0.99669 e^{- 3.8 t}$ .

$0.00331 (- {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} \frac{d}{d t} [0.00331 + 0.99669 e^{- 3.8 t}])$

Step 3

Differentiate.

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Step 3.1

Multiply

$- 1$ by

$0.00331$ .

$- 0.00331 ({(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} \frac{d}{d t} [0.00331 + 0.99669 e^{- 3.8 t}])$

Step 3.2

By the Sum Rule, the derivative of

$0.00331 + 0.99669 e^{- 3.8 t}$ with respect to

$t$ is

$\frac{d}{d t} [0.00331] + \frac{d}{d t} [0.99669 e^{- 3.8 t}]$ .

$- 0.00331 {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} (\frac{d}{d t} [0.00331] + \frac{d}{d t} [0.99669 e^{- 3.8 t}])$

Step 3.3

Since

$0.00331$ is constant with respect to

$t$ , the derivative of

$0.00331$ with respect to

$t$ is

$0$ .

$- 0.00331 {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} (0 + \frac{d}{d t} [0.99669 e^{- 3.8 t}])$

Step 3.4

Add

$0$ and

$\frac{d}{d t} [0.99669 e^{- 3.8 t}]$ .

$- 0.00331 {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} \frac{d}{d t} [0.99669 e^{- 3.8 t}]$

Step 3.5

Since

$0.99669$ is constant with respect to

$t$ , the derivative of

$0.99669 e^{- 3.8 t}$ with respect to

$t$ is

$0.99669 \frac{d}{d t} [e^{- 3.8 t}]$ .

$- 0.00331 {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} (0.99669 \frac{d}{d t} [e^{- 3.8 t}])$

Step 3.6

Multiply

$0.99669$ by

$- 0.00331$ .

$- 0.00329904 {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} \frac{d}{d t} [e^{- 3.8 t}]$

Step 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

$g (t) = - 3.8 t$ .

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Step 4.1

To apply the Chain Rule, set

$u_{2}$ as

$- 3.8 t$ .

$- 0.00329904 {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d t} [- 3.8 t])$

Step 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

$a$ =

$e$ .

$- 0.00329904 {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} (e^{u_{2}} \frac{d}{d t} [- 3.8 t])$

Step 4.3

Replace all occurrences of

$u_{2}$ with

$- 3.8 t$ .

$- 0.00329904 {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} (e^{- 3.8 t} \frac{d}{d t} [- 3.8 t])$

Step 5

Differentiate.

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Step 5.1

Since

$- 3.8$ is constant with respect to

$t$ , the derivative of

$- 3.8 t$ with respect to

$t$ is

$- 3.8 \frac{d}{d t} [t]$ .

$- 0.00329904 {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} (e^{- 3.8 t} (- 3.8 \frac{d}{d t} [t]))$

Step 5.2

Multiply

$- 3.8$ by

$- 0.00329904$ .

$0.01253636 {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} (e^{- 3.8 t} (\frac{d}{d t} [t]))$

Step 5.3

Differentiate using the Power Rule which states that

$\frac{d}{d t} [t^{n}]$ is

$n t^{n - 1}$ where

$n = 1$ .

$0.01253636 {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} (e^{- 3.8 t} \cdot 1)$

Step 5.4

Simplify the expression.

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Step 5.4.1

Multiply

$e^{- 3.8 t}$ by

$1$ .

$0.01253636 {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} e^{- 3.8 t}$

Step 5.4.2

Reorder the factors of

$0.01253636 {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2} e^{- 3.8 t}$ .

$0.01253636 e^{- 3.8 t} {(0.00331 + 0.99669 e^{- 3.8 t})}^{- 2}$