Calculus Examples

$e^{4 x} + e^{- x}$

Step 1

Write $e^{4 x} + e^{- x}$ as a function.

$f (x) = e^{4 x} + e^{- x}$

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

By the Sum Rule, the derivative of $e^{4 x} + e^{- x}$ with respect to $x$ is $\frac{d}{d x} [e^{4 x}] + \frac{d}{d x} [e^{- x}]$ .

$\frac{d}{d x} [e^{4 x}] + \frac{d}{d x} [e^{- x}]$

Step 2.1.2

Evaluate $\frac{d}{d x} [e^{4 x}]$ .

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Step 2.1.2.1

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 $g (x) = 4 x$ .

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Step 2.1.2.1.1

To apply the Chain Rule, set $u_{1}$ as $4 x$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [4 x] + \frac{d}{d x} [e^{- x}]$

Step 2.1.2.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{u_{1}} \frac{d}{d x} [4 x] + \frac{d}{d x} [e^{- x}]$

Step 2.1.2.1.3

Replace all occurrences of $u_{1}$ with $4 x$ .

$e^{4 x} \frac{d}{d x} [4 x] + \frac{d}{d x} [e^{- x}]$

Step 2.1.2.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$e^{4 x} (4 \frac{d}{d x} [x]) + \frac{d}{d x} [e^{- x}]$

Step 2.1.2.3

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

$e^{4 x} (4 \cdot 1) + \frac{d}{d x} [e^{- x}]$

Step 2.1.2.4

Multiply $4$ by $1$ .

$e^{4 x} \cdot 4 + \frac{d}{d x} [e^{- x}]$

Step 2.1.2.5

Move $4$ to the left of $e^{4 x}$ .

$4 e^{4 x} + \frac{d}{d x} [e^{- x}]$

Step 2.1.3

Evaluate $\frac{d}{d x} [e^{- x}]$ .

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Step 2.1.3.1

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 $g (x) = - x$ .

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Step 2.1.3.1.1

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

$4 e^{4 x} + \frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- x]$

Step 2.1.3.1.2

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

$4 e^{4 x} + e^{u_{2}} \frac{d}{d x} [- x]$

Step 2.1.3.1.3

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

$4 e^{4 x} + e^{- x} \frac{d}{d x} [- x]$

Step 2.1.3.2

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

$4 e^{4 x} + e^{- x} (- \frac{d}{d x} [x])$

Step 2.1.3.3

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

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

Step 2.1.3.4

Multiply $- 1$ by $1$ .

$4 e^{4 x} + e^{- x} \cdot - 1$

Step 2.1.3.5

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

$4 e^{4 x} - 1 \cdot e^{- x}$

Step 2.1.3.6

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

$f' (x) = 4 e^{4 x} - e^{- x}$

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $4 e^{4 x} - e^{- x}$ .

$4 e^{4 x} - e^{- x}$

Step 3

Set the first derivative equal to $0$ then solve the equation $4 e^{4 x} - e^{- x} = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

$4 e^{4 x} - e^{- x} = 0$

Step 3.2

Move $- e^{- x}$ to the right side of the equation by adding it to both sides.

$4 e^{4 x} = e^{- x}$

Step 3.3

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

$\ln (4 e^{4 x}) = \ln (e^{- x})$

Step 3.4

Expand the left side.

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Step 3.4.1

Rewrite $\ln (4 e^{4 x})$ as $\ln (4) + \ln (e^{4 x})$ .

$\ln (4) + \ln (e^{4 x}) = \ln (e^{- x})$

Step 3.4.2

Expand $\ln (e^{4 x})$ by moving $4 x$ outside the logarithm.

$\ln (4) + 4 x \ln (e) = \ln (e^{- x})$

Step 3.4.3

The natural logarithm of $e$ is $1$ .

$\ln (4) + 4 x \cdot 1 = \ln (e^{- x})$

Step 3.4.4

Multiply $4$ by $1$ .

$\ln (4) + 4 x = \ln (e^{- x})$

Step 3.5

Expand the right side.

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Step 3.5.1

Expand $\ln (e^{- x})$ by moving $- x$ outside the logarithm.

$\ln (4) + 4 x = - x \ln (e)$

Step 3.5.2

The natural logarithm of $e$ is $1$ .

$\ln (4) + 4 x = - x \cdot 1$

Step 3.5.3

Multiply $- 1$ by $1$ .

$\ln (4) + 4 x = - x$

Step 3.6

Move all terms containing $x$ to the left side of the equation.

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Step 3.6.1

Add $x$ to both sides of the equation.

$\ln (4) + 4 x + x = 0$

Step 3.6.2

Add $4 x$ and $x$ .

$\ln (4) + 5 x = 0$

Step 3.7

Subtract $\ln (4)$ from both sides of the equation.

$5 x = - \ln (4)$

Step 3.8

Divide each term in $5 x = - \ln (4)$ by $5$ and simplify.

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Step 3.8.1

Divide each term in $5 x = - \ln (4)$ by $5$ .

$\frac{5 x}{5} = \frac{- \ln (4)}{5}$

Step 3.8.2

Simplify the left side.

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Step 3.8.2.1

Cancel the common factor of $5$ .

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Step 3.8.2.1.1

Cancel the common factor.

$\frac{5 x}{5} = \frac{- \ln (4)}{5}$

Step 3.8.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \ln (4)}{5}$

Step 3.8.3

Simplify the right side.

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Step 3.8.3.1

Move the negative in front of the fraction.

$x = - \frac{\ln (4)}{5}$

Step 4

The values which make the derivative equal to $0$ are $- \frac{\ln (4)}{5}$ .

$- \frac{\ln (4)}{5}$

Step 5

After finding the point that makes the derivative $f' (x) = 4 e^{4 x} - e^{- x}$ equal to $0$ or undefined, the interval to check where $f (x) = e^{4 x} + e^{- x}$ is increasing and where it is decreasing is $(- \infty, - \frac{\ln (4)}{5}) \cup (- \frac{\ln (4)}{5}, \infty)$ .

$(- \infty, - \frac{\ln (4)}{5}) \cup (- \frac{\ln (4)}{5}, \infty)$

Step 6

Substitute a value from the interval $(- \infty, - \frac{\ln (4)}{5})$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 1.27725887$ in the expression.

$f' (- 1.27725887) = 4 e^{4 (- 1.27725887)} - e^{- (- 1.27725887)}$

Step 6.2

Simplify the result.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Multiply $4$ by $- 1.27725887$ .

$f' (- 1.27725887) = 4 e^{- 5.10903548} - e^{- (- 1.27725887)}$

Step 6.2.1.2

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

$f' (- 1.27725887) = 4 (\frac{1}{e^{5.10903548}}) - e^{- (- 1.27725887)}$

Step 6.2.1.3

Combine $4$ and $\frac{1}{e^{5.10903548}}$ .

$f' (- 1.27725887) = \frac{4}{e^{5.10903548}} - e^{- (- 1.27725887)}$

Step 6.2.1.4

Multiply $- 1$ by $- 1.27725887$ .

$f' (- 1.27725887) = \frac{4}{e^{5.10903548}} - e^{1.27725887}$

Step 6.2.2

The final answer is $\frac{4}{e^{5.10903548}} - e^{1.27725887}$ .

$\frac{4}{e^{5.10903548}} - e^{1.27725887}$

Step 6.3

Simplify.

$- 3.56262674$

Step 6.4

At $x = - 1.27725887$ the derivative is $- 3.56262674$ . Since this is negative, the function is decreasing on $(- \infty, - \frac{\ln (4)}{5})$ .

Decreasing on $(- \infty, - \frac{\ln (4)}{5})$ since $f' (x) < 0$

Step 7

Substitute a value from the interval $(- \frac{\ln (4)}{5}, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $0.72274112$ in the expression.

$f' (0.72274112) = 4 e^{4 (0.72274112)} - e^{- (0.72274112)}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

Multiply $4$ by $0.72274112$ .

$f' (0.72274112) = 4 e^{2.89096451} - e^{- (0.72274112)}$

Step 7.2.1.2

Multiply $- 1$ by $0.72274112$ .

$f' (0.72274112) = 4 e^{2.89096451} - e^{- 0.72274112}$

Step 7.2.1.3

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

$f' (0.72274112) = 4 e^{2.89096451} - \frac{1}{e^{0.72274112}}$

Step 7.2.2

The final answer is $4 e^{2.89096451} - \frac{1}{e^{0.72274112}}$ .

$4 e^{2.89096451} - \frac{1}{e^{0.72274112}}$

Step 7.3

Simplify.

$71.55727104$

Step 7.4

At $x = 0.72274112$ the derivative is $71.55727104$ . Since this is positive, the function is increasing on $(- \frac{\ln (4)}{5}, \infty)$ .

Increasing on $(- \frac{\ln (4)}{5}, \infty)$ since $f' (x) > 0$

Step 8

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- \frac{\ln (4)}{5}, \infty)$

Decreasing on: $(- \infty, - \frac{\ln (4)}{5})$

Step 9