Calculus Examples

$y = \frac{e^{2 x}}{1 + e^{- 2 x}}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = e^{2 x}$ and $g (x) = 1 + e^{- 2 x}$ .

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

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 2 x$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

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

Step 2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 2.3

Replace all occurrences of $u_{1}$ with $2 x$ .

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

Step 3

Differentiate.

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Step 3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 3.2

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

$\frac{(1 + e^{- 2 x}) e^{2 x} (2 \cdot 1) - e^{2 x} \frac{d}{d x} [1 + e^{- 2 x}]}{{(1 + e^{- 2 x})}^{2}}$

Step 3.3

Simplify the expression.

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Step 3.3.1

Multiply $2$ by $1$ .

$\frac{(1 + e^{- 2 x}) e^{2 x} \cdot 2 - e^{2 x} \frac{d}{d x} [1 + e^{- 2 x}]}{{(1 + e^{- 2 x})}^{2}}$

Step 3.3.2

Move $2$ to the left of $(1 + e^{- 2 x}) e^{2 x}$ .

$\frac{2 \cdot ((1 + e^{- 2 x}) e^{2 x}) - e^{2 x} \frac{d}{d x} [1 + e^{- 2 x}]}{{(1 + e^{- 2 x})}^{2}}$

Step 3.4

By the Sum Rule, the derivative of $1 + e^{- 2 x}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [e^{- 2 x}]$ .

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

Step 3.5

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

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

Step 3.6

Add $0$ and $\frac{d}{d x} [e^{- 2 x}]$ .

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

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 $g (x) = - 2 x$ .

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Step 4.1

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

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

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 $a$ = $e$ .

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

Step 4.3

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

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

Step 5

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 6

Simplify by adding terms.

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Step 6.1

Add $- 2 x$ and $2 x$ .

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

Step 6.2

Simplify the expression.

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Step 6.2.1

Anything raised to $0$ is $1$ .

$\frac{2 (1 + e^{- 2 x}) e^{2 x} - 1 \cdot 1 \frac{d}{d x} [- 2 x]}{{(1 + e^{- 2 x})}^{2}}$

Step 6.2.2

Multiply $- 1$ by $1$ .

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

Step 7

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

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

Step 8

Multiply $- 2$ by $- 1$ .

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

Step 9

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

$\frac{2 (1 + e^{- 2 x}) e^{2 x} + 2 \cdot 1}{{(1 + e^{- 2 x})}^{2}}$

Step 10

Multiply $2$ by $1$ .

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

Step 11

Simplify.

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Step 11.1

Apply the distributive property.

$\frac{(2 \cdot 1 + 2 e^{- 2 x}) e^{2 x} + 2}{{(1 + e^{- 2 x})}^{2}}$

Step 11.2

Apply the distributive property.

$\frac{2 \cdot 1 e^{2 x} + 2 e^{- 2 x} e^{2 x} + 2}{{(1 + e^{- 2 x})}^{2}}$

Step 11.3

Simplify the numerator.

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Step 11.3.1

Simplify each term.

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Step 11.3.1.1

Multiply $2$ by $1$ .

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

Step 11.3.1.2

Multiply $e^{- 2 x}$ by $e^{2 x}$ by adding the exponents.

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Step 11.3.1.2.1

Move $e^{2 x}$ .

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

Step 11.3.1.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 11.3.1.2.3

Subtract $2 x$ from $2 x$ .

$\frac{2 e^{2 x} + 2 e^{0} + 2}{{(1 + e^{- 2 x})}^{2}}$

Step 11.3.1.3

Simplify $2 e^{0}$ .

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

Step 11.3.2

Add $2$ and $2$ .

$\frac{2 e^{2 x} + 4}{{(1 + e^{- 2 x})}^{2}}$

Step 11.4

Factor $2$ out of $2 e^{2 x} + 4$ .

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Step 11.4.1

Factor $2$ out of $2 e^{2 x}$ .

$\frac{2 (e^{2 x}) + 4}{{(1 + e^{- 2 x})}^{2}}$

Step 11.4.2

Factor $2$ out of $4$ .

$\frac{2 e^{2 x} + 2 \cdot 2}{{(1 + e^{- 2 x})}^{2}}$

Step 11.4.3

Factor $2$ out of $2 e^{2 x} + 2 \cdot 2$ .

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