Calculus Examples

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

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 $f (x) = e^{2 x} + 1$ and $g (x) = e^{2 x} - 1$ .

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

Step 2

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

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

Step 3

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 3.1

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

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

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 $a$ = $e$ .

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

Step 3.3

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

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

Step 4

Differentiate.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Simplify the expression.

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Step 4.3.1

Multiply $2$ by $1$ .

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

Step 4.3.2

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

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

Step 4.4

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

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

Step 4.5

Simplify the expression.

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Step 4.5.1

Add $2 e^{2 x}$ and $0$ .

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

Step 4.5.2

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

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

Step 4.6

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

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

Step 5

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 5.1

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

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

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 $a$ = $e$ .

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

Step 5.3

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

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

Step 6

Differentiate.

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Step 6.1

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

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

Step 6.2

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

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

Step 6.3

Simplify the expression.

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Step 6.3.1

Multiply $2$ by $1$ .

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

Step 6.3.2

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

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

Step 6.4

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

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

Step 6.5

Simplify the expression.

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Step 6.5.1

Add $2 e^{2 x}$ and $0$ .

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

Step 6.5.2

Multiply $2$ by $- 1$ .

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

Step 7

Simplify.

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Step 7.1

Apply the distributive property.

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

Step 7.2

Apply the distributive property.

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

Step 7.3

Apply the distributive property.

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

Step 7.4

Apply the distributive property.

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

Step 7.5

Simplify the numerator.

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Step 7.5.1

Combine the opposite terms in $2 e^{2 x} e^{2 x} + 2 \cdot - 1 e^{2 x} - 2 e^{2 x} e^{2 x} - 2 \cdot 1 e^{2 x}$ .

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Step 7.5.1.1

Subtract $2 e^{2 x} e^{2 x}$ from $2 e^{2 x} e^{2 x}$ .

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

Step 7.5.1.2

Add $2 \cdot - 1 e^{2 x}$ and $0$ .

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

Step 7.5.2

Simplify each term.

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Step 7.5.2.1

Multiply $2$ by $- 1$ .

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

Step 7.5.2.2

Multiply $- 2$ by $1$ .

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

Step 7.5.3

Subtract $2 e^{2 x}$ from $- 2 e^{2 x}$ .

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

Step 7.6

Move the negative in front of the fraction.

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

Step 7.7

Simplify the denominator.

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Step 7.7.1

Rewrite $e^{2 x}$ as ${(e^{x})}^{2}$ .

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

Step 7.7.2

Rewrite $1$ as $1^{2}$ .

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

Step 7.7.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = e^{x}$ and $b = 1$ .

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

Step 7.7.4

Apply the product rule to $(e^{x} + 1) (e^{x} - 1)$ .

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