Calculus Examples

$f (x) = e^{x} \cos (x)$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = e^{x}$ and $g (x) = \cos (x)$ .

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

Step 1.2

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

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

Step 1.3

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

$e^{x} (- \sin (x)) + \cos (x) e^{x}$

Step 1.4

Reorder terms.

$- e^{x} \sin (x) + e^{x} \cos (x)$

Step 2

Find the second derivative of the function.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = e^{x}$ and $g (x) = \sin (x)$ .

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

Step 2.2.3

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

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

Step 2.2.4

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

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

Step 2.3

Evaluate $\frac{d}{d x} [e^{x} \cos (x)]$ .

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Step 2.3.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = e^{x}$ and $g (x) = \cos (x)$ .

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

Step 2.3.2

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

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

Step 2.3.3

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

$f'' (x) = - (e^{x} \cos (x) + \sin (x) e^{x}) + e^{x} (- \sin (x)) + \cos (x) e^{x}$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$f'' (x) = - (e^{x} \cos (x)) - (\sin (x) e^{x}) + e^{x} (- \sin (x)) + \cos (x) e^{x}$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Reorder $e^{x}$ and $- 1$ .

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

Step 2.4.2.2

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

$f'' (x) = - e^{x} \cos (x) - e^{x} \sin (x) - e^{x} \sin (x) + \cos (x) e^{x}$

Step 2.4.2.3

Subtract $e^{x} \sin (x)$ from $- e^{x} \sin (x)$ .

$f'' (x) = - e^{x} \cos (x) - 2 e^{x} \sin (x) + \cos (x) e^{x}$

Step 2.4.2.4

Add $- e^{x} \cos (x)$ and $\cos (x) e^{x}$ .

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Step 2.4.2.4.1

Reorder $\cos (x)$ and $e^{x}$ .

$f'' (x) = - 2 e^{x} \sin (x) - e^{x} \cos (x) + e^{x} \cos (x)$

Step 2.4.2.4.2

Add $- e^{x} \cos (x)$ and $e^{x} \cos (x)$ .

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

Step 2.4.2.5

Add $- 2 e^{x} \sin (x)$ and $0$ .

$f'' (x) = - 2 e^{x} \sin (x)$

Step 3

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

$- e^{x} \sin (x) + e^{x} \cos (x) = 0$

Step 4

Factor $- e^{x} \sin (x) + e^{x} \cos (x)$ .

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Step 4.1

Factor $e^{x}$ out of $- e^{x} \sin (x) + e^{x} \cos (x)$ .

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Step 4.1.1

Factor $e^{x}$ out of $- e^{x} \sin (x)$ .

$e^{x} (- 1 \sin (x)) + e^{x} \cos (x) = 0$

Step 4.1.2

Factor $e^{x}$ out of $e^{x} \cos (x)$ .

$e^{x} (- 1 \sin (x)) + e^{x} \cos (x) = 0$

Step 4.1.3

Factor $e^{x}$ out of $e^{x} (- 1 \sin (x)) + e^{x} \cos (x)$ .

$e^{x} (- 1 \sin (x) + \cos (x)) = 0$

Step 4.2

Rewrite $- 1 \sin (x)$ as $- \sin (x)$ .

$e^{x} (- \sin (x) + \cos (x)) = 0$

Step 5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$e^{x} = 0$

$- \sin (x) + \cos (x) = 0$

Step 6

Set $e^{x}$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $e^{x}$ equal to $0$ .

$e^{x} = 0$

Step 6.2

Solve $e^{x} = 0$ for $x$ .

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Step 6.2.1

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

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

Step 6.2.2

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

Undefined

Step 6.2.3

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

No solution

Step 7

Set $- \sin (x) + \cos (x)$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $- \sin (x) + \cos (x)$ equal to $0$ .

$- \sin (x) + \cos (x) = 0$

Step 7.2

Solve $- \sin (x) + \cos (x) = 0$ for $x$ .

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Step 7.2.1

Divide each term in the equation by $\cos (x)$ .

$\frac{- \sin (x)}{\cos (x)} + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 7.2.2

Separate fractions.

$\frac{- 1}{1} \cdot \frac{\sin (x)}{\cos (x)} + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 7.2.3

Convert from $\frac{\sin (x)}{\cos (x)}$ to $\tan (x)$ .

$\frac{- 1}{1} \cdot \tan (x) + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 7.2.4

Divide $- 1$ by $1$ .

$- \tan (x) + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 7.2.5

Cancel the common factor of $\cos (x)$ .

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Step 7.2.5.1

Cancel the common factor.

$- \tan (x) + \frac{\cos (x)}{\cos (x)} = \frac{0}{\cos (x)}$

Step 7.2.5.2

Rewrite the expression.

$- \tan (x) + 1 = \frac{0}{\cos (x)}$

Step 7.2.6

Separate fractions.

$- \tan (x) + 1 = \frac{0}{1} \cdot \frac{1}{\cos (x)}$

Step 7.2.7

Convert from $\frac{1}{\cos (x)}$ to $\sec (x)$ .

$- \tan (x) + 1 = \frac{0}{1} \cdot \sec (x)$

Step 7.2.8

Divide $0$ by $1$ .

$- \tan (x) + 1 = 0 \sec (x)$

Step 7.2.9

Multiply $0$ by $\sec (x)$ .

$- \tan (x) + 1 = 0$

Step 7.2.10

Subtract $1$ from both sides of the equation.

$- \tan (x) = - 1$

Step 7.2.11

Divide each term in $- \tan (x) = - 1$ by $- 1$ and simplify.

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Step 7.2.11.1

Divide each term in $- \tan (x) = - 1$ by $- 1$ .

$\frac{- \tan (x)}{- 1} = \frac{- 1}{- 1}$

Step 7.2.11.2

Simplify the left side.

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Step 7.2.11.2.1

Dividing two negative values results in a positive value.

$\frac{\tan (x)}{1} = \frac{- 1}{- 1}$

Step 7.2.11.2.2

Divide $\tan (x)$ by $1$ .

$\tan (x) = \frac{- 1}{- 1}$

Step 7.2.11.3

Simplify the right side.

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Step 7.2.11.3.1

Divide $- 1$ by $- 1$ .

$\tan (x) = 1$

Step 7.2.12

Take the inverse tangent of both sides of the equation to extract $x$ from inside the tangent.

$x = \arctan (1)$

Step 7.2.13

Simplify the right side.

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Step 7.2.13.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$x = \frac{π}{4}$

Step 7.2.14

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x = π + \frac{π}{4}$

Step 7.2.15

Simplify $π + \frac{π}{4}$ .

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Step 7.2.15.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x = π \cdot \frac{4}{4} + \frac{π}{4}$

Step 7.2.15.2

Combine fractions.

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Step 7.2.15.2.1

Combine $π$ and $\frac{4}{4}$ .

$x = \frac{π \cdot 4}{4} + \frac{π}{4}$

Step 7.2.15.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 + π}{4}$

Step 7.2.15.3

Simplify the numerator.

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Step 7.2.15.3.1

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π + π}{4}$

Step 7.2.15.3.2

Add $4 π$ and $π$ .

$x = \frac{5 π}{4}$

Step 7.2.16

The solution to the equation $x = \frac{π}{4}$ .

$x = \frac{π}{4}, \frac{5 π}{4}$

Step 8

The final solution is all the values that make $e^{x} (- \sin (x) + \cos (x)) = 0$ true.

$x = \frac{π}{4}, \frac{5 π}{4}$

Step 9

Evaluate the second derivative at $x = \frac{π}{4}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 2 e^{\frac{π}{4}} \sin (\frac{π}{4})$

Step 10

Evaluate the second derivative.

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Step 10.1

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$- 2 e^{\frac{π}{4}} \frac{\sqrt{2}}{2}$

Step 10.2

Cancel the common factor of $2$ .

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Step 10.2.1

Factor $2$ out of $- 2 e^{\frac{π}{4}}$ .

$2 (- e^{\frac{π}{4}}) \frac{\sqrt{2}}{2}$

Step 10.2.2

Cancel the common factor.

$2 (- e^{\frac{π}{4}}) \frac{\sqrt{2}}{2}$

Step 10.2.3

Rewrite the expression.

$- e^{\frac{π}{4}} \sqrt{2}$

Step 11

$x = \frac{π}{4}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{π}{4}$ is a local maximum

Step 12

Find the y-value when $x = \frac{π}{4}$ .

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Step 12.1

Replace the variable $x$ with $\frac{π}{4}$ in the expression.

$f (\frac{π}{4}) = e^{\frac{π}{4}} \cos (\frac{π}{4})$

Step 12.2

Simplify the result.

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Step 12.2.1

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$f (\frac{π}{4}) = e^{\frac{π}{4}} (\frac{\sqrt{2}}{2})$

Step 12.2.2

Combine $e^{\frac{π}{4}}$ and $\frac{\sqrt{2}}{2}$ .

$f (\frac{π}{4}) = \frac{e^{\frac{π}{4}} \sqrt{2}}{2}$

Step 12.2.3

The final answer is $\frac{e^{\frac{π}{4}} \sqrt{2}}{2}$ .

$y = \frac{e^{\frac{π}{4}} \sqrt{2}}{2}$

Step 13

Evaluate the second derivative at $x = \frac{5 π}{4}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 2 e^{\frac{5 π}{4}} \sin (\frac{5 π}{4})$

Step 14

Evaluate the second derivative.

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Step 14.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the third quadrant.

$- 2 e^{\frac{5 π}{4}} (- \sin (\frac{π}{4}))$

Step 14.2

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$- 2 e^{\frac{5 π}{4}} (- \frac{\sqrt{2}}{2})$

Step 14.3

Cancel the common factor of $2$ .

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Step 14.3.1

Move the leading negative in $- \frac{\sqrt{2}}{2}$ into the numerator.

$- 2 e^{\frac{5 π}{4}} \frac{- \sqrt{2}}{2}$

Step 14.3.2

Factor $2$ out of $- 2 e^{\frac{5 π}{4}}$ .

$2 (- e^{\frac{5 π}{4}}) \frac{- \sqrt{2}}{2}$

Step 14.3.3

Cancel the common factor.

$2 (- e^{\frac{5 π}{4}}) \frac{- \sqrt{2}}{2}$

Step 14.3.4

Rewrite the expression.

$- e^{\frac{5 π}{4}} (- \sqrt{2})$

Step 14.4

Multiply.

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Step 14.4.1

Multiply $- 1$ by $- 1$ .

$1 e^{\frac{5 π}{4}} \sqrt{2}$

Step 14.4.2

Multiply $e^{\frac{5 π}{4}}$ by $1$ .

$e^{\frac{5 π}{4}} \sqrt{2}$

Step 15

$x = \frac{5 π}{4}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{5 π}{4}$ is a local minimum

Step 16

Find the y-value when $x = \frac{5 π}{4}$ .

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Step 16.1

Replace the variable $x$ with $\frac{5 π}{4}$ in the expression.

$f (\frac{5 π}{4}) = e^{\frac{5 π}{4}} \cos (\frac{5 π}{4})$

Step 16.2

Simplify the result.

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Step 16.2.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the third quadrant.

$f (\frac{5 π}{4}) = e^{\frac{5 π}{4}} (- \cos (\frac{π}{4}))$

Step 16.2.2

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$f (\frac{5 π}{4}) = e^{\frac{5 π}{4}} (- \frac{\sqrt{2}}{2})$

Step 16.2.3

Combine $e^{\frac{5 π}{4}}$ and $\frac{\sqrt{2}}{2}$ .

$f (\frac{5 π}{4}) = - \frac{e^{\frac{5 π}{4}} \sqrt{2}}{2}$

Step 16.2.4

The final answer is $- \frac{e^{\frac{5 π}{4}} \sqrt{2}}{2}$ .

$y = - \frac{e^{\frac{5 π}{4}} \sqrt{2}}{2}$

Step 17

These are the local extrema for $f (x) = e^{x} \cos (x)$ .

$(\frac{π}{4}, \frac{e^{\frac{π}{4}} \sqrt{2}}{2})$ is a local maxima

$(\frac{5 π}{4}, - \frac{e^{\frac{5 π}{4}} \sqrt{2}}{2})$ is a local minima

Step 18