Calculus Examples

$f (x) = 5 e^{x} + 5 e^{- x}$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $5 e^{x} + 5 e^{- x}$ with respect to $x$ is $\frac{d}{d x} [5 e^{x}] + \frac{d}{d x} [5 e^{- x}]$ .

$\frac{d}{d x} [5 e^{x}] + \frac{d}{d x} [5 e^{- x}]$

Step 1.2

Evaluate $\frac{d}{d x} [5 e^{x}]$ .

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Step 1.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 e^{x}$ with respect to $x$ is $5 \frac{d}{d x} [e^{x}]$ .

$5 \frac{d}{d x} [e^{x}] + \frac{d}{d x} [5 e^{- x}]$

Step 1.2.2

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

$5 e^{x} + \frac{d}{d x} [5 e^{- x}]$

Step 1.3

Evaluate $\frac{d}{d x} [5 e^{- x}]$ .

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Step 1.3.1

Since $5$ is constant with respect to $x$ , the derivative of $5 e^{- x}$ with respect to $x$ is $5 \frac{d}{d x} [e^{- x}]$ .

$5 e^{x} + 5 \frac{d}{d x} [e^{- x}]$

Step 1.3.2

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

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Step 1.3.2.1

To apply the Chain Rule, set $u$ as $- x$ .

$5 e^{x} + 5 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x])$

Step 1.3.2.2

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

$5 e^{x} + 5 (e^{u} \frac{d}{d x} [- x])$

Step 1.3.2.3

Replace all occurrences of $u$ with $- x$ .

$5 e^{x} + 5 (e^{- x} \frac{d}{d x} [- x])$

Step 1.3.3

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

$5 e^{x} + 5 (e^{- x} (- \frac{d}{d x} [x]))$

Step 1.3.4

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

$5 e^{x} + 5 (e^{- x} (- 1 \cdot 1))$

Step 1.3.5

Multiply $- 1$ by $1$ .

$5 e^{x} + 5 (e^{- x} \cdot - 1)$

Step 1.3.6

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

$5 e^{x} + 5 (- 1 \cdot e^{- x})$

Step 1.3.7

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

$5 e^{x} + 5 (- e^{- x})$

Step 1.3.8

Multiply $- 1$ by $5$ .

$5 e^{x} - 5 e^{- x}$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $5 e^{x} - 5 e^{- x}$ with respect to $x$ is $\frac{d}{d x} [5 e^{x}] + \frac{d}{d x} [- 5 e^{- x}]$ .

$f'' (x) = \frac{d}{d x} (5 e^{x}) + \frac{d}{d x} (- 5 e^{- x})$

Step 2.2

Evaluate $\frac{d}{d x} [5 e^{x}]$ .

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Step 2.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 e^{x}$ with respect to $x$ is $5 \frac{d}{d x} [e^{x}]$ .

$f'' (x) = 5 \frac{d}{d x} (e^{x}) + \frac{d}{d x} (- 5 e^{- x})$

Step 2.2.2

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

$f'' (x) = 5 e^{x} + \frac{d}{d x} (- 5 e^{- x})$

Step 2.3

Evaluate $\frac{d}{d x} [- 5 e^{- x}]$ .

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Step 2.3.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 e^{- x}$ with respect to $x$ is $- 5 \frac{d}{d x} [e^{- x}]$ .

$f'' (x) = 5 e^{x} - 5 \frac{d}{d x} e^{- x}$

Step 2.3.2

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

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Step 2.3.2.1

To apply the Chain Rule, set $u$ as $- x$ .

$f'' (x) = 5 e^{x} - 5 (\frac{d}{d u} (e^{u}) \frac{d}{d x} (- x))$

Step 2.3.2.2

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

$f'' (x) = 5 e^{x} - 5 (e^{u} \frac{d}{d x} (- x))$

Step 2.3.2.3

Replace all occurrences of $u$ with $- x$ .

$f'' (x) = 5 e^{x} - 5 (e^{- x} \frac{d}{d x} (- x))$

Step 2.3.3

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

$f'' (x) = 5 e^{x} - 5 (e^{- x} (- \frac{d}{d x} x))$

Step 2.3.4

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

$f'' (x) = 5 e^{x} - 5 (e^{- x} (- 1 \cdot 1))$

Step 2.3.5

Multiply $- 1$ by $1$ .

$f'' (x) = 5 e^{x} - 5 (e^{- x} \cdot - 1)$

Step 2.3.6

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

$f'' (x) = 5 e^{x} - 5 (- 1 \cdot e^{- x})$

Step 2.3.7

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

$f'' (x) = 5 e^{x} - 5 (- e^{- x})$

Step 2.3.8

Multiply $- 1$ by $- 5$ .

$f'' (x) = 5 e^{x} + 5 e^{- x}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$5 e^{x} - 5 e^{- x} = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

By the Sum Rule, the derivative of $5 e^{x} + 5 e^{- x}$ with respect to $x$ is $\frac{d}{d x} [5 e^{x}] + \frac{d}{d x} [5 e^{- x}]$ .

$\frac{d}{d x} [5 e^{x}] + \frac{d}{d x} [5 e^{- x}]$

Step 4.1.2

Evaluate $\frac{d}{d x} [5 e^{x}]$ .

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Step 4.1.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 e^{x}$ with respect to $x$ is $5 \frac{d}{d x} [e^{x}]$ .

$5 \frac{d}{d x} [e^{x}] + \frac{d}{d x} [5 e^{- x}]$

Step 4.1.2.2

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

$5 e^{x} + \frac{d}{d x} [5 e^{- x}]$

Step 4.1.3

Evaluate $\frac{d}{d x} [5 e^{- x}]$ .

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Step 4.1.3.1

Since $5$ is constant with respect to $x$ , the derivative of $5 e^{- x}$ with respect to $x$ is $5 \frac{d}{d x} [e^{- x}]$ .

$5 e^{x} + 5 \frac{d}{d x} [e^{- x}]$

Step 4.1.3.2

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

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Step 4.1.3.2.1

To apply the Chain Rule, set $u$ as $- x$ .

$5 e^{x} + 5 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x])$

Step 4.1.3.2.2

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

$5 e^{x} + 5 (e^{u} \frac{d}{d x} [- x])$

Step 4.1.3.2.3

Replace all occurrences of $u$ with $- x$ .

$5 e^{x} + 5 (e^{- x} \frac{d}{d x} [- x])$

Step 4.1.3.3

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

$5 e^{x} + 5 (e^{- x} (- \frac{d}{d x} [x]))$

Step 4.1.3.4

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

$5 e^{x} + 5 (e^{- x} (- 1 \cdot 1))$

Step 4.1.3.5

Multiply $- 1$ by $1$ .

$5 e^{x} + 5 (e^{- x} \cdot - 1)$

Step 4.1.3.6

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

$5 e^{x} + 5 (- 1 \cdot e^{- x})$

Step 4.1.3.7

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

$5 e^{x} + 5 (- e^{- x})$

Step 4.1.3.8

Multiply $- 1$ by $5$ .

$f' (x) = 5 e^{x} - 5 e^{- x}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $5 e^{x} - 5 e^{- x}$ .

$5 e^{x} - 5 e^{- x}$

Step 5

Set the first derivative equal to $0$ then solve the equation $5 e^{x} - 5 e^{- x} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$5 e^{x} - 5 e^{- x} = 0$

Step 5.2

Move $- 5 e^{- x}$ to the right side of the equation by adding it to both sides.

$5 e^{x} = 5 e^{- x}$

Step 5.3

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

$\ln (5 e^{x}) = \ln (5 e^{- x})$

Step 5.4

Expand the left side.

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Step 5.4.1

Rewrite $\ln (5 e^{x})$ as $\ln (5) + \ln (e^{x})$ .

$\ln (5) + \ln (e^{x}) = \ln (5 e^{- x})$

Step 5.4.2

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

$\ln (5) + x \ln (e) = \ln (5 e^{- x})$

Step 5.4.3

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

$\ln (5) + x \cdot 1 = \ln (5 e^{- x})$

Step 5.4.4

Multiply $x$ by $1$ .

$\ln (5) + x = \ln (5 e^{- x})$

Step 5.5

Expand the right side.

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Step 5.5.1

Rewrite $\ln (5 e^{- x})$ as $\ln (5) + \ln (e^{- x})$ .

$\ln (5) + x = \ln (5) + \ln (e^{- x})$

Step 5.5.2

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

$\ln (5) + x = \ln (5) - x \ln (e)$

Step 5.5.3

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

$\ln (5) + x = \ln (5) - x \cdot 1$

Step 5.5.4

Multiply $- 1$ by $1$ .

$\ln (5) + x = \ln (5) - x$

Step 5.6

Move all the terms containing a logarithm to the left side of the equation.

$\ln (5) - \ln (5) = - x - x$

Step 5.7

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

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

Step 5.8

Divide $5$ by $5$ .

$\ln (1) = - x - x$

Step 5.9

The natural logarithm of $1$ is $0$ .

$0 = - x - x$

Step 5.10

Subtract $x$ from $- x$ .

$0 = - 2 x$

Step 5.11

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- 2 x = 0$

Step 5.12

Divide each term in $- 2 x = 0$ by $- 2$ and simplify.

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Step 5.12.1

Divide each term in $- 2 x = 0$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{0}{- 2}$

Step 5.12.2

Simplify the left side.

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Step 5.12.2.1

Cancel the common factor of $- 2$ .

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Step 5.12.2.1.1

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{0}{- 2}$

Step 5.12.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{- 2}$

Step 5.12.3

Simplify the right side.

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Step 5.12.3.1

Divide $0$ by $- 2$ .

$x = 0$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = 0$

Step 8

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$5 e^{0} + 5 e^{- (0)}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

Anything raised to $0$ is $1$ .

$5 \cdot 1 + 5 e^{- (0)}$

Step 9.1.2

Multiply $5$ by $1$ .

$5 + 5 e^{- (0)}$

Step 9.1.3

Multiply $- 1$ by $0$ .

$5 + 5 e^{0}$

Step 9.1.4

Anything raised to $0$ is $1$ .

$5 + 5 \cdot 1$

Step 9.1.5

Multiply $5$ by $1$ .

$5 + 5$

Step 9.2

Add $5$ and $5$ .

$10$

Step 10

$x = 0$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 0$ is a local minimum

Step 11

Find the y-value when $x = 0$ .

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Step 11.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = 5 e^{0} + 5 e^{- (0)}$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Anything raised to $0$ is $1$ .

$f (0) = 5 \cdot 1 + 5 e^{- (0)}$

Step 11.2.1.2

Multiply $5$ by $1$ .

$f (0) = 5 + 5 e^{- (0)}$

Step 11.2.1.3

Multiply $- 1$ by $0$ .

$f (0) = 5 + 5 e^{0}$

Step 11.2.1.4

Anything raised to $0$ is $1$ .

$f (0) = 5 + 5 \cdot 1$

Step 11.2.1.5

Multiply $5$ by $1$ .

$f (0) = 5 + 5$

Step 11.2.2

Add $5$ and $5$ .

$f (0) = 10$

Step 11.2.3

The final answer is $10$ .

$y = 10$

Step 12

These are the local extrema for $f (x) = 5 e^{x} + 5 e^{- x}$ .

$(0, 10)$ is a local minima

Step 13