Calculus Examples

$y = \frac{500}{12 + 4 e^{- 0.5 x}}$

Step 1

Differentiate using the Constant Multiple Rule.

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Step 1.1

Cancel the common factor of $500$ and $12 + 4 e^{- 0.5 x}$ .

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Step 1.1.1

Factor $4$ out of $500$ .

$\frac{d}{d x} [\frac{4 \cdot 125}{12 + 4 e^{- 0.5 x}}]$

Step 1.1.2

Cancel the common factors.

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Step 1.1.2.1

Factor $4$ out of $12$ .

$\frac{d}{d x} [\frac{4 \cdot 125}{4 (3) + 4 e^{- 0.5 x}}]$

Step 1.1.2.2

Factor $4$ out of $4 e^{- 0.5 x}$ .

$\frac{d}{d x} [\frac{4 \cdot 125}{4 (3) + 4 (e^{- 0.5 x})}]$

Step 1.1.2.3

Factor $4$ out of $4 (3) + 4 (e^{- 0.5 x})$ .

$\frac{d}{d x} [\frac{4 \cdot 125}{4 (3 + e^{- 0.5 x})}]$

Step 1.1.2.4

Cancel the common factor.

$\frac{d}{d x} [\frac{4 \cdot 125}{4 (3 + e^{- 0.5 x})}]$

Step 1.1.2.5

Rewrite the expression.

$\frac{d}{d x} [\frac{125}{3 + e^{- 0.5 x}}]$

Step 1.2

Since $125$ is constant with respect to $x$ , the derivative of $\frac{125}{3 + e^{- 0.5 x}}$ with respect to $x$ is $125 \frac{d}{d x} [\frac{1}{3 + e^{- 0.5 x}}]$ .

$125 \frac{d}{d x} [\frac{1}{3 + e^{- 0.5 x}}]$

Step 1.3

Rewrite $\frac{1}{3 + e^{- 0.5 x}}$ as ${(3 + e^{- 0.5 x})}^{- 1}$ .

$125 \frac{d}{d x} [{(3 + e^{- 0.5 x})}^{- 1}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = 3 + e^{- 0.5 x}$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $3 + e^{- 0.5 x}$ .

$125 (\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d x} [3 + e^{- 0.5 x}])$

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 $n = - 1$ .

$125 (- {u_{1}}^{- 2} \frac{d}{d x} [3 + e^{- 0.5 x}])$

Step 2.3

Replace all occurrences of $u_{1}$ with $3 + e^{- 0.5 x}$ .

$125 (- {(3 + e^{- 0.5 x})}^{- 2} \frac{d}{d x} [3 + e^{- 0.5 x}])$

Step 3

Differentiate.

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Step 3.1

Multiply $- 1$ by $125$ .

$- 125 ({(3 + e^{- 0.5 x})}^{- 2} \frac{d}{d x} [3 + e^{- 0.5 x}])$

Step 3.2

By the Sum Rule, the derivative of $3 + e^{- 0.5 x}$ with respect to $x$ is $\frac{d}{d x} [3] + \frac{d}{d x} [e^{- 0.5 x}]$ .

$- 125 {(3 + e^{- 0.5 x})}^{- 2} (\frac{d}{d x} [3] + \frac{d}{d x} [e^{- 0.5 x}])$

Step 3.3

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$- 125 {(3 + e^{- 0.5 x})}^{- 2} (0 + \frac{d}{d x} [e^{- 0.5 x}])$

Step 3.4

Add $0$ and $\frac{d}{d x} [e^{- 0.5 x}]$ .

$- 125 {(3 + e^{- 0.5 x})}^{- 2} \frac{d}{d x} [e^{- 0.5 x}]$

Step 4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 0.5 x$ .

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $- 0.5 x$ .

$- 125 {(3 + e^{- 0.5 x})}^{- 2} (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- 0.5 x])$

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 $a$ = $e$ .

$- 125 {(3 + e^{- 0.5 x})}^{- 2} (e^{u_{2}} \frac{d}{d x} [- 0.5 x])$

Step 4.3

Replace all occurrences of $u_{2}$ with $- 0.5 x$ .

$- 125 {(3 + e^{- 0.5 x})}^{- 2} (e^{- 0.5 x} \frac{d}{d x} [- 0.5 x])$

Step 5

Differentiate.

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Step 5.1

Since $- 0.5$ is constant with respect to $x$ , the derivative of $- 0.5 x$ with respect to $x$ is $- 0.5 \frac{d}{d x} [x]$ .

$- 125 {(3 + e^{- 0.5 x})}^{- 2} (e^{- 0.5 x} (- 0.5 \frac{d}{d x} [x]))$

Step 5.2

Multiply $- 0.5$ by $- 125$ .

$62.5 {(3 + e^{- 0.5 x})}^{- 2} (e^{- 0.5 x} (\frac{d}{d x} [x]))$

Step 5.3

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

$62.5 {(3 + e^{- 0.5 x})}^{- 2} (e^{- 0.5 x} \cdot 1)$

Step 5.4

Multiply $e^{- 0.5 x}$ by $1$ .

$62.5 {(3 + e^{- 0.5 x})}^{- 2} e^{- 0.5 x}$

Step 6

Simplify.

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Step 6.1

Reorder the factors of $62.5 {(3 + e^{- 0.5 x})}^{- 2} e^{- 0.5 x}$ .

$62.5 e^{- 0.5 x} {(3 + e^{- 0.5 x})}^{- 2}$

Step 6.2

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

$62.5 e^{- 0.5 x} \frac{1}{{(3 + e^{- 0.5 x})}^{2}}$

Step 6.3

Multiply $62.5 e^{- 0.5 x} \frac{1}{{(3 + e^{- 0.5 x})}^{2}}$ .

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Step 6.3.1

Combine $\frac{1}{{(3 + e^{- 0.5 x})}^{2}}$ and $62.5$ .

$\frac{62.5}{{(3 + e^{- 0.5 x})}^{2}} e^{- 0.5 x}$

Step 6.3.2

Combine $\frac{62.5}{{(3 + e^{- 0.5 x})}^{2}}$ and $e^{- 0.5 x}$ .

$\frac{62.5 e^{- 0.5 x}}{{(3 + e^{- 0.5 x})}^{2}}$