Calculus Examples

$\frac{e^{x}}{3 + e^{x}}$

Step 1

Write $\frac{e^{x}}{3 + e^{x}}$ as a function.

$f (x) = \frac{e^{x}}{3 + e^{x}}$

Step 2

Find the first derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

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 $f (x) = e^{x}$ and $g (x) = 3 + e^{x}$ .

$f' (x) = \frac{(3 + e^{x}) \frac{d}{d x} (e^{x}) - e^{x} \frac{d}{d x} (3 + e^{x})}{{(3 + e^{x})}^{2}}$

Step 2.1.2

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

$f' (x) = \frac{(3 + e^{x}) e^{x} - e^{x} \frac{d}{d x} (3 + e^{x})}{{(3 + e^{x})}^{2}}$

Step 2.1.3

Differentiate.

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Step 2.1.3.1

By the Sum Rule, the derivative of $3 + e^{x}$ with respect to $x$ is $\frac{d}{d x} [3] + \frac{d}{d x} [e^{x}]$ .

$f' (x) = \frac{(3 + e^{x}) e^{x} - e^{x} (\frac{d}{d x} (3) + \frac{d}{d x} (e^{x}))}{{(3 + e^{x})}^{2}}$

Step 2.1.3.2

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$f' (x) = \frac{(3 + e^{x}) e^{x} - e^{x} (0 + \frac{d}{d x} (e^{x}))}{{(3 + e^{x})}^{2}}$

Step 2.1.3.3

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

$f' (x) = \frac{(3 + e^{x}) e^{x} - e^{x} \frac{d}{d x} e^{x}}{{(3 + e^{x})}^{2}}$

Step 2.1.4

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

$f' (x) = \frac{(3 + e^{x}) e^{x} - e^{x} e^{x}}{{(3 + e^{x})}^{2}}$

Step 2.1.5

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

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Step 2.1.5.1

Move $e^{x}$ .

$f' (x) = \frac{(3 + e^{x}) e^{x} - (e^{x} e^{x})}{{(3 + e^{x})}^{2}}$

Step 2.1.5.2

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

$f' (x) = \frac{(3 + e^{x}) e^{x} - e^{x + x}}{{(3 + e^{x})}^{2}}$

Step 2.1.5.3

Add $x$ and $x$ .

$f' (x) = \frac{(3 + e^{x}) e^{x} - e^{2 x}}{{(3 + e^{x})}^{2}}$

Step 2.1.6

Simplify.

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Step 2.1.6.1

Apply the distributive property.

$f' (x) = \frac{3 e^{x} + e^{x} e^{x} - e^{2 x}}{{(3 + e^{x})}^{2}}$

Step 2.1.6.2

Simplify the numerator.

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Step 2.1.6.2.1

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

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Step 2.1.6.2.1.1

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

$f' (x) = \frac{3 e^{x} + e^{x + x} - e^{2 x}}{{(3 + e^{x})}^{2}}$

Step 2.1.6.2.1.2

Add $x$ and $x$ .

$f' (x) = \frac{3 e^{x} + e^{2 x} - e^{2 x}}{{(3 + e^{x})}^{2}}$

Step 2.1.6.2.2

Combine the opposite terms in $3 e^{x} + e^{2 x} - e^{2 x}$ .

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Step 2.1.6.2.2.1

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

$f' (x) = \frac{3 e^{x} + 0}{{(3 + e^{x})}^{2}}$

Step 2.1.6.2.2.2

Add $3 e^{x}$ and $0$ .

$f' (x) = \frac{3 e^{x}}{{(3 + e^{x})}^{2}}$

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{3 e^{x}}{{(3 + e^{x})}^{2}}$ .

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

Step 3

Set the first derivative equal to $0$ then solve the equation $\frac{3 e^{x}}{{(3 + e^{x})}^{2}} = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

$\frac{3 e^{x}}{{(3 + e^{x})}^{2}} = 0$

Step 3.2

Set the numerator equal to zero.

$3 e^{x} = 0$

Step 3.3

Solve the equation for $x$ .

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Step 3.3.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (3 e^{x}) = \ln (0)$

Step 3.3.2

The equation cannot be solved because $\ln (0)$ is undefined.

Undefined

Step 3.3.3

There is no solution for $3 e^{x} = 0$

No solution

Step 4

There are no values of $x$ in the domain of the original problem where the derivative is $0$ or undefined.

No critical points found

Step 5

No points make the derivative $f' (x) = \frac{3 e^{x}}{{(3 + e^{x})}^{2}}$ equal to $0$ or undefined. The interval to check if $f (x) = \frac{e^{x}}{3 + e^{x}}$ is increasing or decreasing is $(- \infty, \infty)$ .

$(- \infty, \infty)$

Step 6

Substitute any number, such as $1$ , from the interval $(- \infty, \infty)$ in the derivative $f' (x) = \frac{3 e^{x}}{{(3 + e^{x})}^{2}}$ to check if the result is negative or positive. If the result is negative, the graph is decreasing on the interval $(- \infty, \infty)$ . If the result is positive, the graph is increasing on the interval $(- \infty, \infty)$ .

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Step 6.1

Replace the variable $x$ with $1$ in the expression.

$f' (1) = \frac{3 e^{1}}{{(3 + e^{1})}^{2}}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify.

$f' (1) = \frac{3 e}{{(3 + e)}^{2}}$

Step 6.2.2

The final answer is $\frac{3 e}{{(3 + e)}^{2}}$ .

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

Step 7

The result of substituting $1$ into $f' (x) = \frac{3 e^{x}}{{(3 + e^{x})}^{2}}$ is $\frac{3 e}{{(3 + e)}^{2}}$ , which is positive, so the graph is increasing on the interval $(- \infty, \infty)$ .

Increasing on $(- \infty, \infty)$ since $\frac{3 e^{x}}{{(3 + e^{x})}^{2}} > 0$

Step 8

Increasing over the interval $(- \infty, \infty)$ means that the function is always increasing.

Always Increasing

Step 9