Calculus Examples

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

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

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

Step 2

Differentiate.

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Step 2.1

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

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

Step 2.2

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

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

Step 3

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

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

Step 4

Differentiate.

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Step 4.1

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

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

Step 4.2

Simplify the expression.

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Step 4.2.1

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

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

Step 4.2.2

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

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

Step 4.3

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

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

Step 4.4

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

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

Step 5

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

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

Step 6

Differentiate using the Constant Rule.

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Step 6.1

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

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

Step 6.2

Simplify the expression.

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Step 6.2.1

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

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

Step 6.2.2

Multiply $2$ by $- 1$ .

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

Step 7

Simplify.

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Step 7.1

Apply the distributive property.

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

Step 7.2

Apply the distributive property.

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

Step 7.3

Apply the distributive property.

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

Step 7.4

Apply the distributive property.

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

Step 7.5

Simplify the numerator.

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Step 7.5.1

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

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Step 7.5.1.1

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

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

Step 7.5.1.2

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

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

Step 7.5.2

Simplify each term.

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Step 7.5.2.1

Multiply $2$ by $1$ .

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

Step 7.5.2.2

Multiply $- 2$ by $- 1$ .

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

Step 7.5.3

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

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