Calculus Examples

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

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 $f (u) = e^{2 u}$ and $g (u) = e^{u} + e^{- u}$ .

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

Step 2

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 2.1

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

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

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 $a$ = $e$ .

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

Step 2.3

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

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

Step 3

Differentiate.

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Simplify the expression.

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Step 3.3.1

Multiply $2$ by $1$ .

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

Step 3.3.2

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

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

Step 3.4

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

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

Step 4

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

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

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 $g (u) = - u$ .

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Step 5.1

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

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

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 $a$ = $e$ .

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

Step 5.3

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

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

Step 6

Differentiate.

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Simplify the expression.

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Step 6.3.1

Multiply $- 1$ by $1$ .

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

Step 6.3.2

Move $- 1$ to the left of $e^{- u}$ .

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

Step 6.3.3

Rewrite $- 1 e^{- u}$ as $- e^{- u}$ .

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

Step 7

Simplify.

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Step 7.1

Apply the distributive property.

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

Step 7.2

Apply the distributive property.

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

Step 7.3

Apply the distributive property.

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

Step 7.4

Simplify the numerator.

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Step 7.4.1

Simplify each term.

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Step 7.4.1.1

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

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Step 7.4.1.1.1

Move $e^{2 u}$ .

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

Step 7.4.1.1.2

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

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

Step 7.4.1.1.3

Add $2 u$ and $u$ .

$\frac{2 e^{3 u} + 2 e^{- u} e^{2 u} - e^{2 u} e^{u} - e^{2 u} (- e^{- u})}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.1.2

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

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Step 7.4.1.2.1

Move $e^{2 u}$ .

$\frac{2 e^{3 u} + 2 (e^{2 u} e^{- u}) - e^{2 u} e^{u} - e^{2 u} (- e^{- u})}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.1.2.2

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

$\frac{2 e^{3 u} + 2 e^{2 u - u} - e^{2 u} e^{u} - e^{2 u} (- e^{- u})}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.1.2.3

Subtract $u$ from $2 u$ .

$\frac{2 e^{3 u} + 2 e^{u} - e^{2 u} e^{u} - e^{2 u} (- e^{- u})}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.1.3

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

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Step 7.4.1.3.1

Move $e^{u}$ .

$\frac{2 e^{3 u} + 2 e^{u} - (e^{u} e^{2 u}) - e^{2 u} (- e^{- u})}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.1.3.2

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

$\frac{2 e^{3 u} + 2 e^{u} - e^{u + 2 u} - e^{2 u} (- e^{- u})}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.1.3.3

Add $u$ and $2 u$ .

$\frac{2 e^{3 u} + 2 e^{u} - e^{3 u} - e^{2 u} (- e^{- u})}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.1.4

Rewrite using the commutative property of multiplication.

$\frac{2 e^{3 u} + 2 e^{u} - e^{3 u} - 1 \cdot - 1 e^{2 u} e^{- u}}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.1.5

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

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Step 7.4.1.5.1

Move $e^{- u}$ .

$\frac{2 e^{3 u} + 2 e^{u} - e^{3 u} - 1 \cdot - 1 (e^{- u} e^{2 u})}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.1.5.2

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

$\frac{2 e^{3 u} + 2 e^{u} - e^{3 u} - 1 \cdot - 1 e^{- u + 2 u}}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.1.5.3

Add $- u$ and $2 u$ .

$\frac{2 e^{3 u} + 2 e^{u} - e^{3 u} - 1 \cdot - 1 e^{u}}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.1.6

Multiply $- 1$ by $- 1$ .

$\frac{2 e^{3 u} + 2 e^{u} - e^{3 u} + 1 e^{u}}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.1.7

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

$\frac{2 e^{3 u} + 2 e^{u} - e^{3 u} + e^{u}}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.2

Subtract $e^{3 u}$ from $2 e^{3 u}$ .

$\frac{2 e^{u} + e^{3 u} + e^{u}}{{(e^{u} + e^{- u})}^{2}}$

Step 7.4.3

Add $2 e^{u}$ and $e^{u}$ .

$\frac{3 e^{u} + e^{3 u}}{{(e^{u} + e^{- u})}^{2}}$