Calculus Examples

$\frac{e^{2 u} - e^{- 2 u}}{e^{2 u} + e^{- 2 u}}$

Step 1

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

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

Step 2

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

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

Step 3

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

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 4

Differentiate.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Simplify the expression.

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Step 4.3.1

Multiply $2$ by $1$ .

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

Step 4.3.2

Move $2$ to the left of $e^{2 u}$ .

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

Step 4.4

Since $- 1$ is constant with respect to $u$ , the derivative of $- e^{- 2 u}$ with respect to $u$ is $- \frac{d}{d u} [e^{- 2 u}]$ .

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

Step 5

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

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Step 5.1

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

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

Step 5.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

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

Step 5.3

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

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

Step 6

Differentiate.

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Step 6.1

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

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

Step 6.2

Multiply $- 2$ by $- 1$ .

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

Step 6.3

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

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

Step 6.4

Multiply $2$ by $1$ .

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

Step 6.5

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

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

Step 7

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

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Step 7.1

To apply the Chain Rule, set $u_{3}$ as $2 u$ .

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

Step 7.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{3}} [a^{u_{3}}]$ is $a^{u_{3}} \ln (a)$ where $a$ = $e$ .

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

Step 7.3

Replace all occurrences of $u_{3}$ with $2 u$ .

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

Step 8

Differentiate.

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Step 8.1

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

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

Step 8.2

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

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

Step 8.3

Simplify the expression.

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Step 8.3.1

Multiply $2$ by $1$ .

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

Step 8.3.2

Move $2$ to the left of $e^{2 u}$ .

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

Step 9

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

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Step 9.1

To apply the Chain Rule, set $u_{4}$ as $- 2 u$ .

$\frac{(e^{2 u} + e^{- 2 u}) (2 e^{2 u} + 2 e^{- 2 u}) - (e^{2 u} - e^{- 2 u}) (2 e^{2 u} + \frac{d}{d u_{4}} [e^{u_{4}}] \frac{d}{d u} [- 2 u])}{{(e^{2 u} + e^{- 2 u})}^{2}}$

Step 9.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{4}} [a^{u_{4}}]$ is $a^{u_{4}} \ln (a)$ where $a$ = $e$ .

$\frac{(e^{2 u} + e^{- 2 u}) (2 e^{2 u} + 2 e^{- 2 u}) - (e^{2 u} - e^{- 2 u}) (2 e^{2 u} + e^{u_{4}} \frac{d}{d u} [- 2 u])}{{(e^{2 u} + e^{- 2 u})}^{2}}$

Step 9.3

Replace all occurrences of $u_{4}$ with $- 2 u$ .

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

Step 10

Differentiate.

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Step 10.1

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

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

Step 10.2

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

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

Step 10.3

Simplify the expression.

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Step 10.3.1

Multiply $- 2$ by $1$ .

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

Step 10.3.2

Move $- 2$ to the left of $e^{- 2 u}$ .

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

Step 11

Simplify.

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Step 11.1

Apply the distributive property.

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

Step 11.2

Simplify the numerator.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Expand $(e^{2 u} + e^{- 2 u}) (2 e^{2 u} + 2 e^{- 2 u})$ using the FOIL Method.

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Step 11.2.1.1.1

Apply the distributive property.

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

Step 11.2.1.1.2

Apply the distributive property.

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

Step 11.2.1.1.3

Apply the distributive property.

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

Step 11.2.1.2

Simplify and combine like terms.

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Step 11.2.1.2.1

Simplify each term.

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Step 11.2.1.2.1.1

Rewrite using the commutative property of multiplication.

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

Step 11.2.1.2.1.2

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

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Step 11.2.1.2.1.2.1

Move $e^{2 u}$ .

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

Step 11.2.1.2.1.2.2

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

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

Step 11.2.1.2.1.2.3

Add $2 u$ and $2 u$ .

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

Step 11.2.1.2.1.3

Rewrite using the commutative property of multiplication.

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

Step 11.2.1.2.1.4

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

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Step 11.2.1.2.1.4.1

Move $e^{- 2 u}$ .

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

Step 11.2.1.2.1.4.2

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

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

Step 11.2.1.2.1.4.3

Add $- 2 u$ and $2 u$ .

$\frac{2 e^{4 u} + 2 e^{0} + e^{- 2 u} (2 e^{2 u}) + e^{- 2 u} (2 e^{- 2 u}) + (- e^{2 u} - - e^{- 2 u}) (2 e^{2 u} - 2 e^{- 2 u})}{{(e^{2 u} + e^{- 2 u})}^{2}}$

Step 11.2.1.2.1.5

Simplify $2 e^{0}$ .

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

Step 11.2.1.2.1.6

Rewrite using the commutative property of multiplication.

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

Step 11.2.1.2.1.7

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

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Step 11.2.1.2.1.7.1

Move $e^{2 u}$ .

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

Step 11.2.1.2.1.7.2

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

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

Step 11.2.1.2.1.7.3

Subtract $2 u$ from $2 u$ .

$\frac{2 e^{4 u} + 2 + 2 e^{0} + e^{- 2 u} (2 e^{- 2 u}) + (- e^{2 u} - - e^{- 2 u}) (2 e^{2 u} - 2 e^{- 2 u})}{{(e^{2 u} + e^{- 2 u})}^{2}}$

Step 11.2.1.2.1.8

Simplify $2 e^{0}$ .

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

Step 11.2.1.2.1.9

Rewrite using the commutative property of multiplication.

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

Step 11.2.1.2.1.10

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

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Step 11.2.1.2.1.10.1

Move $e^{- 2 u}$ .

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

Step 11.2.1.2.1.10.2

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

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

Step 11.2.1.2.1.10.3

Subtract $2 u$ from $- 2 u$ .

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

Step 11.2.1.2.2

Add $2$ and $2$ .

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

Step 11.2.1.3

Multiply $- - e^{- 2 u}$ .

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Step 11.2.1.3.1

Multiply $- 1$ by $- 1$ .

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

Step 11.2.1.3.2

Multiply $e^{- 2 u}$ by $1$ .

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

Step 11.2.1.4

Expand $(- e^{2 u} + e^{- 2 u}) (2 e^{2 u} - 2 e^{- 2 u})$ using the FOIL Method.

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Step 11.2.1.4.1

Apply the distributive property.

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

Step 11.2.1.4.2

Apply the distributive property.

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

Step 11.2.1.4.3

Apply the distributive property.

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

Step 11.2.1.5

Simplify and combine like terms.

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Step 11.2.1.5.1

Simplify each term.

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Step 11.2.1.5.1.1

Rewrite using the commutative property of multiplication.

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

Step 11.2.1.5.1.2

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

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Step 11.2.1.5.1.2.1

Move $e^{2 u}$ .

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

Step 11.2.1.5.1.2.2

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

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

Step 11.2.1.5.1.2.3

Add $2 u$ and $2 u$ .

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

Step 11.2.1.5.1.3

Multiply $- 1$ by $2$ .

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

Step 11.2.1.5.1.4

Rewrite using the commutative property of multiplication.

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

Step 11.2.1.5.1.5

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

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Step 11.2.1.5.1.5.1

Move $e^{- 2 u}$ .

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

Step 11.2.1.5.1.5.2

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

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

Step 11.2.1.5.1.5.3

Add $- 2 u$ and $2 u$ .

$\frac{2 e^{4 u} + 4 + 2 e^{- 4 u} - 2 e^{4 u} - 1 \cdot - 2 e^{0} + e^{- 2 u} (2 e^{2 u}) + e^{- 2 u} (- 2 e^{- 2 u})}{{(e^{2 u} + e^{- 2 u})}^{2}}$

Step 11.2.1.5.1.6

Simplify $- 1 \cdot - 2 e^{0}$ .

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

Step 11.2.1.5.1.7

Multiply $- 1$ by $- 2$ .

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

Step 11.2.1.5.1.8

Rewrite using the commutative property of multiplication.

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

Step 11.2.1.5.1.9

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

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Step 11.2.1.5.1.9.1

Move $e^{2 u}$ .

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

Step 11.2.1.5.1.9.2

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

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

Step 11.2.1.5.1.9.3

Subtract $2 u$ from $2 u$ .

$\frac{2 e^{4 u} + 4 + 2 e^{- 4 u} - 2 e^{4 u} + 2 + 2 e^{0} + e^{- 2 u} (- 2 e^{- 2 u})}{{(e^{2 u} + e^{- 2 u})}^{2}}$

Step 11.2.1.5.1.10

Simplify $2 e^{0}$ .

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

Step 11.2.1.5.1.11

Rewrite using the commutative property of multiplication.

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

Step 11.2.1.5.1.12

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

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Step 11.2.1.5.1.12.1

Move $e^{- 2 u}$ .

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

Step 11.2.1.5.1.12.2

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

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

Step 11.2.1.5.1.12.3

Subtract $2 u$ from $- 2 u$ .

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

Step 11.2.1.5.2

Add $2$ and $2$ .

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

Step 11.2.2

Combine the opposite terms in $2 e^{4 u} + 4 + 2 e^{- 4 u} - 2 e^{4 u} + 4 - 2 e^{- 4 u}$ .

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Step 11.2.2.1

Subtract $2 e^{4 u}$ from $2 e^{4 u}$ .

$\frac{4 + 2 e^{- 4 u} + 0 + 4 - 2 e^{- 4 u}}{{(e^{2 u} + e^{- 2 u})}^{2}}$

Step 11.2.2.2

Add $4 + 2 e^{- 4 u}$ and $0$ .

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

Step 11.2.2.3

Subtract $2 e^{- 4 u}$ from $2 e^{- 4 u}$ .

$\frac{4 + 0 + 4}{{(e^{2 u} + e^{- 2 u})}^{2}}$

Step 11.2.2.4

Add $4$ and $0$ .

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

Step 11.2.3

Add $4$ and $4$ .

$\frac{8}{{(e^{2 u} + e^{- 2 u})}^{2}}$