Calculus Examples

$20 {(1 + e^{10 - t})}^{- 1}$

Step 1

Since $20$ is constant with respect to $t$ , the derivative of $20 {(1 + e^{10 - t})}^{- 1}$ with respect to $t$ is $20 \frac{d}{d t} [{(1 + e^{10 - t})}^{- 1}]$ .

$20 \frac{d}{d t} [{(1 + e^{10 - t})}^{- 1}]$

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 $g (t) = 1 + e^{10 - t}$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $1 + e^{10 - t}$ .

$20 (\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d t} [1 + e^{10 - t}])$

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 $n = - 1$ .

$20 (- {u_{1}}^{- 2} \frac{d}{d t} [1 + e^{10 - t}])$

Step 2.3

Replace all occurrences of $u_{1}$ with $1 + e^{10 - t}$ .

$20 (- {(1 + e^{10 - t})}^{- 2} \frac{d}{d t} [1 + e^{10 - t}])$

Step 3

Differentiate.

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Step 3.1

Multiply $- 1$ by $20$ .

$- 20 ({(1 + e^{10 - t})}^{- 2} \frac{d}{d t} [1 + e^{10 - t}])$

Step 3.2

By the Sum Rule, the derivative of $1 + e^{10 - t}$ with respect to $t$ is $\frac{d}{d t} [1] + \frac{d}{d t} [e^{10 - t}]$ .

$- 20 {(1 + e^{10 - t})}^{- 2} (\frac{d}{d t} [1] + \frac{d}{d t} [e^{10 - t}])$

Step 3.3

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

$- 20 {(1 + e^{10 - t})}^{- 2} (0 + \frac{d}{d t} [e^{10 - t}])$

Step 3.4

Add $0$ and $\frac{d}{d t} [e^{10 - t}]$ .

$- 20 {(1 + e^{10 - t})}^{- 2} \frac{d}{d t} [e^{10 - t}]$

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 $g (t) = 10 - t$ .

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $10 - t$ .

$- 20 {(1 + e^{10 - t})}^{- 2} (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d t} [10 - t])$

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 $a$ = $e$ .

$- 20 {(1 + e^{10 - t})}^{- 2} (e^{u_{2}} \frac{d}{d t} [10 - t])$

Step 4.3

Replace all occurrences of $u_{2}$ with $10 - t$ .

$- 20 {(1 + e^{10 - t})}^{- 2} (e^{10 - t} \frac{d}{d t} [10 - t])$

Step 5

Differentiate.

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Step 5.1

By the Sum Rule, the derivative of $10 - t$ with respect to $t$ is $\frac{d}{d t} [10] + \frac{d}{d t} [- t]$ .

$- 20 {(1 + e^{10 - t})}^{- 2} (e^{10 - t} (\frac{d}{d t} [10] + \frac{d}{d t} [- t]))$

Step 5.2

Since $10$ is constant with respect to $t$ , the derivative of $10$ with respect to $t$ is $0$ .

$- 20 {(1 + e^{10 - t})}^{- 2} (e^{10 - t} (0 + \frac{d}{d t} [- t]))$

Step 5.3

Add $0$ and $\frac{d}{d t} [- t]$ .

$- 20 {(1 + e^{10 - t})}^{- 2} (e^{10 - t} \frac{d}{d t} [- t])$

Step 5.4

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

$- 20 {(1 + e^{10 - t})}^{- 2} (e^{10 - t} (- \frac{d}{d t} [t]))$

Step 5.5

Multiply $- 1$ by $- 20$ .

$20 {(1 + e^{10 - t})}^{- 2} (e^{10 - t} (\frac{d}{d t} [t]))$

Step 5.6

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

$20 {(1 + e^{10 - t})}^{- 2} (e^{10 - t} \cdot 1)$

Step 5.7

Multiply $e^{10 - t}$ by $1$ .

$20 {(1 + e^{10 - t})}^{- 2} e^{10 - t}$

Step 6

Simplify.

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Step 6.1

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

$20 \frac{1}{{(1 + e^{10 - t})}^{2}} e^{10 - t}$

Step 6.2

Combine terms.

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Step 6.2.1

Combine $20$ and $\frac{1}{{(1 + e^{10 - t})}^{2}}$ .

$\frac{20}{{(1 + e^{10 - t})}^{2}} e^{10 - t}$

Step 6.2.2

Combine $\frac{20}{{(1 + e^{10 - t})}^{2}}$ and $e^{10 - t}$ .

$\frac{20 e^{10 - t}}{{(1 + e^{10 - t})}^{2}}$