Calculus Examples

$f (x) = \frac{20}{1 + 9 e^{- 3 x}}$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Differentiate using the Constant Multiple Rule.

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Step 1.1.1.1

Since $20$ is constant with respect to $x$ , the derivative of $\frac{20}{1 + 9 e^{- 3 x}}$ with respect to $x$ is $20 \frac{d}{d x} [\frac{1}{1 + 9 e^{- 3 x}}]$ .

$20 \frac{d}{d x} [\frac{1}{1 + 9 e^{- 3 x}}]$

Step 1.1.1.2

Rewrite $\frac{1}{1 + 9 e^{- 3 x}}$ as ${(1 + 9 e^{- 3 x})}^{- 1}$ .

$20 \frac{d}{d x} [{(1 + 9 e^{- 3 x})}^{- 1}]$

Step 1.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = 1 + 9 e^{- 3 x}$ .

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Step 1.1.2.1

To apply the Chain Rule, set $u_{1}$ as $1 + 9 e^{- 3 x}$ .

$20 (\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d x} [1 + 9 e^{- 3 x}])$

Step 1.1.2.2

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

$20 (- {u_{1}}^{- 2} \frac{d}{d x} [1 + 9 e^{- 3 x}])$

Step 1.1.2.3

Replace all occurrences of $u_{1}$ with $1 + 9 e^{- 3 x}$ .

$20 (- {(1 + 9 e^{- 3 x})}^{- 2} \frac{d}{d x} [1 + 9 e^{- 3 x}])$

Step 1.1.3

Differentiate.

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Step 1.1.3.1

Multiply $- 1$ by $20$ .

$- 20 ({(1 + 9 e^{- 3 x})}^{- 2} \frac{d}{d x} [1 + 9 e^{- 3 x}])$

Step 1.1.3.2

By the Sum Rule, the derivative of $1 + 9 e^{- 3 x}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [9 e^{- 3 x}]$ .

$- 20 {(1 + 9 e^{- 3 x})}^{- 2} (\frac{d}{d x} [1] + \frac{d}{d x} [9 e^{- 3 x}])$

Step 1.1.3.3

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

$- 20 {(1 + 9 e^{- 3 x})}^{- 2} (0 + \frac{d}{d x} [9 e^{- 3 x}])$

Step 1.1.3.4

Add $0$ and $\frac{d}{d x} [9 e^{- 3 x}]$ .

$- 20 {(1 + 9 e^{- 3 x})}^{- 2} \frac{d}{d x} [9 e^{- 3 x}]$

Step 1.1.3.5

Since $9$ is constant with respect to $x$ , the derivative of $9 e^{- 3 x}$ with respect to $x$ is $9 \frac{d}{d x} [e^{- 3 x}]$ .

$- 20 {(1 + 9 e^{- 3 x})}^{- 2} (9 \frac{d}{d x} [e^{- 3 x}])$

Step 1.1.3.6

Multiply $9$ by $- 20$ .

$- 180 {(1 + 9 e^{- 3 x})}^{- 2} \frac{d}{d x} [e^{- 3 x}]$

Step 1.1.4

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 $g (x) = - 3 x$ .

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Step 1.1.4.1

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

$- 180 {(1 + 9 e^{- 3 x})}^{- 2} (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- 3 x])$

Step 1.1.4.2

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

$- 180 {(1 + 9 e^{- 3 x})}^{- 2} (e^{u_{2}} \frac{d}{d x} [- 3 x])$

Step 1.1.4.3

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

$- 180 {(1 + 9 e^{- 3 x})}^{- 2} (e^{- 3 x} \frac{d}{d x} [- 3 x])$

Step 1.1.5

Differentiate.

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Step 1.1.5.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x$ with respect to $x$ is $- 3 \frac{d}{d x} [x]$ .

$- 180 {(1 + 9 e^{- 3 x})}^{- 2} (e^{- 3 x} (- 3 \frac{d}{d x} [x]))$

Step 1.1.5.2

Multiply $- 3$ by $- 180$ .

$540 {(1 + 9 e^{- 3 x})}^{- 2} (e^{- 3 x} (\frac{d}{d x} [x]))$

Step 1.1.5.3

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

$540 {(1 + 9 e^{- 3 x})}^{- 2} (e^{- 3 x} \cdot 1)$

Step 1.1.5.4

Multiply $e^{- 3 x}$ by $1$ .

$540 {(1 + 9 e^{- 3 x})}^{- 2} e^{- 3 x}$

Step 1.1.6

Simplify.

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Step 1.1.6.1

Reorder the factors of $540 {(1 + 9 e^{- 3 x})}^{- 2} e^{- 3 x}$ .

$540 e^{- 3 x} {(1 + 9 e^{- 3 x})}^{- 2}$

Step 1.1.6.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$540 e^{- 3 x} \frac{1}{{(1 + 9 e^{- 3 x})}^{2}}$

Step 1.1.6.3

Multiply $540 e^{- 3 x} \frac{1}{{(1 + 9 e^{- 3 x})}^{2}}$ .

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Step 1.1.6.3.1

Combine $\frac{1}{{(1 + 9 e^{- 3 x})}^{2}}$ and $540$ .

$\frac{540}{{(1 + 9 e^{- 3 x})}^{2}} e^{- 3 x}$

Step 1.1.6.3.2

Combine $\frac{540}{{(1 + 9 e^{- 3 x})}^{2}}$ and $e^{- 3 x}$ .

$f' (x) = \frac{540 e^{- 3 x}}{{(1 + 9 e^{- 3 x})}^{2}}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{540 e^{- 3 x}}{{(1 + 9 e^{- 3 x})}^{2}}$ .

$\frac{540 e^{- 3 x}}{{(1 + 9 e^{- 3 x})}^{2}}$

Step 2

Set the first derivative equal to $0$ then solve the equation $\frac{540 e^{- 3 x}}{{(1 + 9 e^{- 3 x})}^{2}} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$\frac{540 e^{- 3 x}}{{(1 + 9 e^{- 3 x})}^{2}} = 0$

Step 2.2

Set the numerator equal to zero.

$540 e^{- 3 x} = 0$

Step 2.3

Solve the equation for $x$ .

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Step 2.3.1

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

$\ln (540 e^{- 3 x}) = \ln (0)$

Step 2.3.2

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

Undefined

Step 2.3.3

There is no solution for $540 e^{- 3 x} = 0$

No solution

Step 3

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 4

No points make the derivative $f' (x) = \frac{540 e^{- 3 x}}{{(1 + 9 e^{- 3 x})}^{2}}$ equal to $0$ or undefined. The interval to check if $f (x) = \frac{20}{1 + 9 e^{- 3 x}}$ is increasing or decreasing is $(- \infty, \infty)$ .

$(- \infty, \infty)$

Step 5

Substitute any number, such as $1$ , from the interval $(- \infty, \infty)$ in the derivative $f' (x) = \frac{540 e^{- 3 x}}{{(1 + 9 e^{- 3 x})}^{2}}$ 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 5.1

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

$f' (1) = \frac{540 e^{- 3 \cdot 1}}{{(1 + 9 e^{- 3 \cdot 1})}^{2}}$

Step 5.2

Simplify the result.

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Step 5.2.1

Move $e^{- 3 \cdot 1}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (1) = \frac{540}{{(1 + 9 e^{- 3})}^{2} e^{3}}$

Step 5.2.2

Simplify the denominator.

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Step 5.2.2.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (1) = \frac{540}{{(1 + 9 (\frac{1}{e^{3}}))}^{2} e^{3}}$

Step 5.2.2.2

Combine $9$ and $\frac{1}{e^{3}}$ .

$f' (1) = \frac{540}{{(1 + \frac{9}{e^{3}})}^{2} e^{3}}$

Step 5.2.2.3

Write $1$ as a fraction with a common denominator.

$f' (1) = \frac{540}{{(\frac{e^{3}}{e^{3}} + \frac{9}{e^{3}})}^{2} e^{3}}$

Step 5.2.2.4

Combine the numerators over the common denominator.

$f' (1) = \frac{540}{{(\frac{e^{3} + 9}{e^{3}})}^{2} e^{3}}$

Step 5.2.2.5

Apply the product rule to $\frac{e^{3} + 9}{e^{3}}$ .

$f' (1) = \frac{540}{\frac{{(e^{3} + 9)}^{2}}{{(e^{3})}^{2}} \cdot e^{3}}$

Step 5.2.2.6

Multiply the exponents in ${(e^{3})}^{2}$ .

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Step 5.2.2.6.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f' (1) = \frac{540}{\frac{{(e^{3} + 9)}^{2}}{e^{3 \cdot 2}} \cdot e^{3}}$

Step 5.2.2.6.2

Multiply $3$ by $2$ .

$f' (1) = \frac{540}{\frac{{(e^{3} + 9)}^{2}}{e^{6}} \cdot e^{3}}$

Step 5.2.3

Simplify terms.

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Step 5.2.3.1

Combine $\frac{{(e^{3} + 9)}^{2}}{e^{6}}$ and $e^{3}$ .

$f' (1) = \frac{540}{\frac{{(e^{3} + 9)}^{2} e^{3}}{e^{6}}}$

Step 5.2.3.2

Reduce the expression $\frac{{(e^{3} + 9)}^{2} e^{3}}{e^{6}}$ by cancelling the common factors.

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Step 5.2.3.2.1

Factor $e^{3}$ out of ${(e^{3} + 9)}^{2} e^{3}$ .

$f' (1) = \frac{540}{\frac{e^{3} {(e^{3} + 9)}^{2}}{e^{6}}}$

Step 5.2.3.2.2

Factor $e^{3}$ out of $e^{6}$ .

$f' (1) = \frac{540}{\frac{e^{3} {(e^{3} + 9)}^{2}}{e^{3} e^{3}}}$

Step 5.2.3.2.3

Cancel the common factor.

$f' (1) = \frac{540}{\frac{e^{3} {(e^{3} + 9)}^{2}}{e^{3} e^{3}}}$

Step 5.2.3.2.4

Rewrite the expression.

$f' (1) = \frac{540}{\frac{{(e^{3} + 9)}^{2}}{e^{3}}}$

Step 5.2.4

Multiply the numerator by the reciprocal of the denominator.

$f' (1) = 540 (\frac{e^{3}}{{(e^{3} + 9)}^{2}})$

Step 5.2.5

Combine $540$ and $\frac{e^{3}}{{(e^{3} + 9)}^{2}}$ .

$f' (1) = \frac{540 e^{3}}{{(e^{3} + 9)}^{2}}$

Step 5.2.6

The final answer is $\frac{540 e^{3}}{{(e^{3} + 9)}^{2}}$ .

$\frac{540 e^{3}}{{(e^{3} + 9)}^{2}}$

Step 6

The result of substituting $1$ into $f' (x) = \frac{540 e^{- 3 x}}{{(1 + 9 e^{- 3 x})}^{2}}$ is $\frac{540 e^{3}}{{(e^{3} + 9)}^{2}}$ , which is positive, so the graph is increasing on the interval $(- \infty, \infty)$ .

Increasing on $(- \infty, \infty)$ since $\frac{540 e^{- 3 x}}{{(1 + 9 e^{- 3 x})}^{2}} > 0$

Step 7

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

Always Increasing

Step 8