Calculus Examples

$\frac{10000}{9 + 4 e^{- 0.2 t}}$

Step 1

Differentiate using the Constant Multiple Rule.

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Step 1.1

Since $10000$ is constant with respect to $t$ , the derivative of $\frac{10000}{9 + 4 e^{- 0.2 t}}$ with respect to $t$ is $10000 \frac{d}{d t} [\frac{1}{9 + 4 e^{- 0.2 t}}]$ .

$10000 \frac{d}{d t} [\frac{1}{9 + 4 e^{- 0.2 t}}]$

Step 1.2

Rewrite $\frac{1}{9 + 4 e^{- 0.2 t}}$ as ${(9 + 4 e^{- 0.2 t})}^{- 1}$ .

$10000 \frac{d}{d t} [{(9 + 4 e^{- 0.2 t})}^{- 1}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{- 1}$ and $g (t) = 9 + 4 e^{- 0.2 t}$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $9 + 4 e^{- 0.2 t}$ .

$10000 (\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d t} [9 + 4 e^{- 0.2 t}])$

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 $n = - 1$ .

$10000 (- {u_{1}}^{- 2} \frac{d}{d t} [9 + 4 e^{- 0.2 t}])$

Step 2.3

Replace all occurrences of $u_{1}$ with $9 + 4 e^{- 0.2 t}$ .

$10000 (- {(9 + 4 e^{- 0.2 t})}^{- 2} \frac{d}{d t} [9 + 4 e^{- 0.2 t}])$

Step 3

Differentiate.

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Step 3.1

Multiply $- 1$ by $10000$ .

$- 10000 ({(9 + 4 e^{- 0.2 t})}^{- 2} \frac{d}{d t} [9 + 4 e^{- 0.2 t}])$

Step 3.2

By the Sum Rule, the derivative of $9 + 4 e^{- 0.2 t}$ with respect to $t$ is $\frac{d}{d t} [9] + \frac{d}{d t} [4 e^{- 0.2 t}]$ .

$- 10000 {(9 + 4 e^{- 0.2 t})}^{- 2} (\frac{d}{d t} [9] + \frac{d}{d t} [4 e^{- 0.2 t}])$

Step 3.3

Since $9$ is constant with respect to $t$ , the derivative of $9$ with respect to $t$ is $0$ .

$- 10000 {(9 + 4 e^{- 0.2 t})}^{- 2} (0 + \frac{d}{d t} [4 e^{- 0.2 t}])$

Step 3.4

Add $0$ and $\frac{d}{d t} [4 e^{- 0.2 t}]$ .

$- 10000 {(9 + 4 e^{- 0.2 t})}^{- 2} \frac{d}{d t} [4 e^{- 0.2 t}]$

Step 3.5

Since $4$ is constant with respect to $t$ , the derivative of $4 e^{- 0.2 t}$ with respect to $t$ is $4 \frac{d}{d t} [e^{- 0.2 t}]$ .

$- 10000 {(9 + 4 e^{- 0.2 t})}^{- 2} (4 \frac{d}{d t} [e^{- 0.2 t}])$

Step 3.6

Multiply $4$ by $- 10000$ .

$- 40000 {(9 + 4 e^{- 0.2 t})}^{- 2} \frac{d}{d t} [e^{- 0.2 t}]$

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 $g (t) = - 0.2 t$ .

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $- 0.2 t$ .

$- 40000 {(9 + 4 e^{- 0.2 t})}^{- 2} (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d t} [- 0.2 t])$

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 $a$ = $e$ .

$- 40000 {(9 + 4 e^{- 0.2 t})}^{- 2} (e^{u_{2}} \frac{d}{d t} [- 0.2 t])$

Step 4.3

Replace all occurrences of $u_{2}$ with $- 0.2 t$ .

$- 40000 {(9 + 4 e^{- 0.2 t})}^{- 2} (e^{- 0.2 t} \frac{d}{d t} [- 0.2 t])$

Step 5

Differentiate.

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Step 5.1

Since $- 0.2$ is constant with respect to $t$ , the derivative of $- 0.2 t$ with respect to $t$ is $- 0.2 \frac{d}{d t} [t]$ .

$- 40000 {(9 + 4 e^{- 0.2 t})}^{- 2} (e^{- 0.2 t} (- 0.2 \frac{d}{d t} [t]))$

Step 5.2

Multiply $- 0.2$ by $- 40000$ .

$8000 {(9 + 4 e^{- 0.2 t})}^{- 2} (e^{- 0.2 t} (\frac{d}{d t} [t]))$

Step 5.3

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

$8000 {(9 + 4 e^{- 0.2 t})}^{- 2} (e^{- 0.2 t} \cdot 1)$

Step 5.4

Multiply $e^{- 0.2 t}$ by $1$ .

$8000 {(9 + 4 e^{- 0.2 t})}^{- 2} e^{- 0.2 t}$

Step 6

Simplify.

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Step 6.1

Reorder the factors of $8000 {(9 + 4 e^{- 0.2 t})}^{- 2} e^{- 0.2 t}$ .

$8000 e^{- 0.2 t} {(9 + 4 e^{- 0.2 t})}^{- 2}$

Step 6.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$8000 e^{- 0.2 t} \frac{1}{{(9 + 4 e^{- 0.2 t})}^{2}}$

Step 6.3

Multiply $8000 e^{- 0.2 t} \frac{1}{{(9 + 4 e^{- 0.2 t})}^{2}}$ .

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Step 6.3.1

Combine $\frac{1}{{(9 + 4 e^{- 0.2 t})}^{2}}$ and $8000$ .

$\frac{8000}{{(9 + 4 e^{- 0.2 t})}^{2}} e^{- 0.2 t}$

Step 6.3.2

Combine $\frac{8000}{{(9 + 4 e^{- 0.2 t})}^{2}}$ and $e^{- 0.2 t}$ .

$\frac{8000 e^{- 0.2 t}}{{(9 + 4 e^{- 0.2 t})}^{2}}$