Calculus Examples

$f (x) = \frac{e^{x} + e^{- x}}{6}$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

Since $\frac{1}{6}$ is constant with respect to $x$ , the derivative of $\frac{e^{x} + e^{- x}}{6}$ with respect to $x$ is $\frac{1}{6} \frac{d}{d x} [e^{x} + e^{- x}]$ .

$\frac{1}{6} \frac{d}{d x} [e^{x} + e^{- x}]$

Step 1.1.2

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

$\frac{1}{6} (\frac{d}{d x} [e^{x}] + \frac{d}{d x} [e^{- x}])$

Step 1.2

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

$\frac{1}{6} (e^{x} + \frac{d}{d x} [e^{- x}])$

Step 1.3

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 1.3.1

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

$\frac{1}{6} (e^{x} + \frac{d}{d u} [e^{u}] \frac{d}{d x} [- x])$

Step 1.3.2

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

$\frac{1}{6} (e^{x} + e^{u} \frac{d}{d x} [- x])$

Step 1.3.3

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

$\frac{1}{6} (e^{x} + e^{- x} \frac{d}{d x} [- x])$

Step 1.4

Differentiate.

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Step 1.4.1

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

$\frac{1}{6} (e^{x} + e^{- x} (- \frac{d}{d x} [x]))$

Step 1.4.2

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

$\frac{1}{6} (e^{x} + e^{- x} (- 1 \cdot 1))$

Step 1.4.3

Simplify the expression.

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Step 1.4.3.1

Multiply $- 1$ by $1$ .

$\frac{1}{6} (e^{x} + e^{- x} \cdot - 1)$

Step 1.4.3.2

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

$\frac{1}{6} (e^{x} - 1 \cdot e^{- x})$

Step 1.4.3.3

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

$\frac{1}{6} (e^{x} - e^{- x})$

Step 1.5

Simplify.

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Step 1.5.1

Apply the distributive property.

$\frac{1}{6} e^{x} + \frac{1}{6} (- e^{- x})$

Step 1.5.2

Combine $\frac{1}{6}$ and $e^{- x}$ .

$\frac{1}{6} e^{x} - \frac{e^{- x}}{6}$

Step 1.5.3

Combine $\frac{1}{6}$ and $e^{x}$ .

$\frac{e^{x}}{6} - \frac{e^{- x}}{6}$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $\frac{e^{x}}{6} - \frac{e^{- x}}{6}$ with respect to $x$ is $\frac{d}{d x} [\frac{e^{x}}{6}] + \frac{d}{d x} [- \frac{e^{- x}}{6}]$ .

$f'' (x) = \frac{d}{d x} (\frac{e^{x}}{6}) + \frac{d}{d x} (- \frac{e^{- x}}{6})$

Step 2.2

Evaluate $\frac{d}{d x} [\frac{e^{x}}{6}]$ .

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Step 2.2.1

Since $\frac{1}{6}$ is constant with respect to $x$ , the derivative of $\frac{e^{x}}{6}$ with respect to $x$ is $\frac{1}{6} \frac{d}{d x} [e^{x}]$ .

$f'' (x) = \frac{1}{6} \cdot \frac{d}{d x} (e^{x}) + \frac{d}{d x} (- \frac{e^{- x}}{6})$

Step 2.2.2

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

$f'' (x) = \frac{1}{6} \cdot e^{x} + \frac{d}{d x} (- \frac{e^{- x}}{6})$

Step 2.3

Evaluate $\frac{d}{d x} [- \frac{e^{- x}}{6}]$ .

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Step 2.3.1

Since $- \frac{1}{6}$ is constant with respect to $x$ , the derivative of $- \frac{e^{- x}}{6}$ with respect to $x$ is $- \frac{1}{6} \frac{d}{d x} [e^{- x}]$ .

$f'' (x) = \frac{1}{6} \cdot e^{x} - \frac{1}{6} \cdot \frac{d}{d x} e^{- x}$

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 $- x$ .

$f'' (x) = \frac{1}{6} \cdot e^{x} - \frac{1}{6} \cdot (\frac{d}{d u} (e^{u}) \frac{d}{d x} (- x))$

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 $a$ = $e$ .

$f'' (x) = \frac{1}{6} \cdot e^{x} - \frac{1}{6} \cdot (e^{u} \frac{d}{d x} (- x))$

Step 2.3.2.3

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

$f'' (x) = \frac{1}{6} \cdot e^{x} - \frac{1}{6} \cdot (e^{- x} \frac{d}{d x} (- x))$

Step 2.3.3

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

$f'' (x) = \frac{1}{6} \cdot e^{x} - \frac{1}{6} \cdot (e^{- x} (- \frac{d}{d x} x))$

Step 2.3.4

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

$f'' (x) = \frac{1}{6} \cdot e^{x} - \frac{1}{6} \cdot (e^{- x} (- 1 \cdot 1))$

Step 2.3.5

Multiply $- 1$ by $1$ .

$f'' (x) = \frac{1}{6} \cdot e^{x} - \frac{1}{6} \cdot (e^{- x} \cdot - 1)$

Step 2.3.6

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

$f'' (x) = \frac{1}{6} \cdot e^{x} - \frac{1}{6} \cdot (- 1 \cdot e^{- x})$

Step 2.3.7

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

$f'' (x) = \frac{1}{6} \cdot e^{x} - \frac{1}{6} \cdot (- e^{- x})$

Step 2.3.8

Multiply $- 1$ by $- 1$ .

$f'' (x) = \frac{1}{6} \cdot e^{x} + 1 (\frac{1}{6}) \cdot e^{- x}$

Step 2.3.9

Multiply $\frac{1}{6}$ by $1$ .

$f'' (x) = \frac{1}{6} \cdot e^{x} + \frac{1}{6} \cdot e^{- x}$

Step 2.3.10

Combine $\frac{1}{6}$ and $e^{- x}$ .

$f'' (x) = \frac{1}{6} \cdot e^{x} + \frac{e^{- x}}{6}$

Step 2.4

Combine $\frac{1}{6}$ and $e^{x}$ .

$f'' (x) = \frac{e^{x}}{6} + \frac{e^{- x}}{6}$

Step 3

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

$\frac{e^{x}}{6} - \frac{e^{- x}}{6} = 0$

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

Differentiate.

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Step 4.1.1.1

Since $\frac{1}{6}$ is constant with respect to $x$ , the derivative of $\frac{e^{x} + e^{- x}}{6}$ with respect to $x$ is $\frac{1}{6} \frac{d}{d x} [e^{x} + e^{- x}]$ .

$\frac{1}{6} \frac{d}{d x} [e^{x} + e^{- x}]$

Step 4.1.1.2

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

$\frac{1}{6} (\frac{d}{d x} [e^{x}] + \frac{d}{d x} [e^{- x}])$

Step 4.1.2

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

$\frac{1}{6} (e^{x} + \frac{d}{d x} [e^{- x}])$

Step 4.1.3

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 4.1.3.1

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

$\frac{1}{6} (e^{x} + \frac{d}{d u} [e^{u}] \frac{d}{d x} [- x])$

Step 4.1.3.2

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

$\frac{1}{6} (e^{x} + e^{u} \frac{d}{d x} [- x])$

Step 4.1.3.3

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

$\frac{1}{6} (e^{x} + e^{- x} \frac{d}{d x} [- x])$

Step 4.1.4

Differentiate.

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Step 4.1.4.1

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

$\frac{1}{6} (e^{x} + e^{- x} (- \frac{d}{d x} [x]))$

Step 4.1.4.2

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

$\frac{1}{6} (e^{x} + e^{- x} (- 1 \cdot 1))$

Step 4.1.4.3

Simplify the expression.

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Step 4.1.4.3.1

Multiply $- 1$ by $1$ .

$\frac{1}{6} (e^{x} + e^{- x} \cdot - 1)$

Step 4.1.4.3.2

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

$\frac{1}{6} (e^{x} - 1 \cdot e^{- x})$

Step 4.1.4.3.3

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

$\frac{1}{6} (e^{x} - e^{- x})$

Step 4.1.5

Simplify.

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Step 4.1.5.1

Apply the distributive property.

$\frac{1}{6} e^{x} + \frac{1}{6} (- e^{- x})$

Step 4.1.5.2

Combine $\frac{1}{6}$ and $e^{- x}$ .

$\frac{1}{6} e^{x} - \frac{e^{- x}}{6}$

Step 4.1.5.3

Combine $\frac{1}{6}$ and $e^{x}$ .

$f' (x) = \frac{e^{x}}{6} - \frac{e^{- x}}{6}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{e^{x}}{6} - \frac{e^{- x}}{6}$ .

$\frac{e^{x}}{6} - \frac{e^{- x}}{6}$

Step 5

Set the first derivative equal to $0$ then solve the equation $\frac{e^{x}}{6} - \frac{e^{- x}}{6} = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$\frac{e^{x}}{6} - \frac{e^{- x}}{6} = 0$

Step 5.2

Move $- \frac{e^{- x}}{6}$ to the right side of the equation by adding it to both sides.

$\frac{e^{x}}{6} = \frac{e^{- x}}{6}$

Step 5.3

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

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

Step 5.4

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$x = - x$

Step 5.5

Solve for $x$ .

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Step 5.5.1

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

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Step 5.5.1.1

Add $x$ to both sides of the equation.

$x + x = 0$

Step 5.5.1.2

Add $x$ and $x$ .

$2 x = 0$

Step 5.5.2

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

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Step 5.5.2.1

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

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

Step 5.5.2.2

Simplify the left side.

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Step 5.5.2.2.1

Cancel the common factor of $2$ .

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Step 5.5.2.2.1.1

Cancel the common factor.

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

Step 5.5.2.2.1.2

Divide $x$ by $1$ .

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

Step 5.5.2.3

Simplify the right side.

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Step 5.5.2.3.1

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.

$\frac{e^{0}}{6} + \frac{e^{- (0)}}{6}$

Step 9

Evaluate the second derivative.

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Step 9.1

Combine the numerators over the common denominator.

$\frac{e^{0} + e^{- (0)}}{6}$

Step 9.2

Simplify each term.

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Step 9.2.1

Anything raised to $0$ is $1$ .

$\frac{1 + e^{- (0)}}{6}$

Step 9.2.2

Multiply $- 1$ by $0$ .

$\frac{1 + e^{0}}{6}$

Step 9.2.3

Anything raised to $0$ is $1$ .

$\frac{1 + 1}{6}$

Step 9.3

Reduce the expression by cancelling the common factors.

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Step 9.3.1

Add $1$ and $1$ .

$\frac{2}{6}$

Step 9.3.2

Cancel the common factor of $2$ and $6$ .

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Step 9.3.2.1

Factor $2$ out of $2$ .

$\frac{2 (1)}{6}$

Step 9.3.2.2

Cancel the common factors.

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Step 9.3.2.2.1

Factor $2$ out of $6$ .

$\frac{2 \cdot 1}{2 \cdot 3}$

Step 9.3.2.2.2

Cancel the common factor.

$\frac{2 \cdot 1}{2 \cdot 3}$

Step 9.3.2.2.3

Rewrite the expression.

$\frac{1}{3}$

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.

$f (0) = \frac{e^{0} + e^{- (0)}}{6}$

Step 11.2

Simplify the result.

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Step 11.2.1

Simplify the numerator.

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Step 11.2.1.1

Anything raised to $0$ is $1$ .

$f (0) = \frac{1 + e^{- (0)}}{6}$

Step 11.2.1.2

Multiply $- 1$ by $0$ .

$f (0) = \frac{1 + e^{0}}{6}$

Step 11.2.1.3

Anything raised to $0$ is $1$ .

$f (0) = \frac{1 + 1}{6}$

Step 11.2.1.4

Add $1$ and $1$ .

$f (0) = \frac{2}{6}$

Step 11.2.2

Cancel the common factor of $2$ and $6$ .

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Step 11.2.2.1

Factor $2$ out of $2$ .

$f (0) = \frac{2 (1)}{6}$

Step 11.2.2.2

Cancel the common factors.

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Step 11.2.2.2.1

Factor $2$ out of $6$ .

$f (0) = \frac{2 \cdot 1}{2 \cdot 3}$

Step 11.2.2.2.2

Cancel the common factor.

$f (0) = \frac{2 \cdot 1}{2 \cdot 3}$

Step 11.2.2.2.3

Rewrite the expression.

$f (0) = \frac{1}{3}$

Step 11.2.3

The final answer is $\frac{1}{3}$ .

$y = \frac{1}{3}$

Step 12

These are the local extrema for $f (x) = \frac{e^{x} + e^{- x}}{6}$ .

$(0, \frac{1}{3})$ is a local minima

Step 13