Calculus Examples

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

Step 1

Find the first derivative.

Tap for more steps...

Step 1.1

Find the first derivative.

Tap for more steps...

Step 1.1.1

Since $2$ is constant with respect to $x$ , the derivative of $2 e^{- x} \cos (x)$ with respect to $x$ is $2 \frac{d}{d x} [e^{- x} \cos (x)]$ .

$2 \frac{d}{d x} [e^{- x} \cos (x)]$

Step 1.1.2

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 $g (x) = \cos (x)$ .

$2 (e^{- x} \frac{d}{d x} [\cos (x)] + \cos (x) \frac{d}{d x} [e^{- x}])$

Step 1.1.3

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

$2 (e^{- x} (- \sin (x)) + \cos (x) \frac{d}{d x} [e^{- x}])$

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 $g (x) = - x$ .

Tap for more steps...

Step 1.1.4.1

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

$2 (e^{- x} (- \sin (x)) + \cos (x) (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x]))$

Step 1.1.4.2

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

$2 (e^{- x} (- \sin (x)) + \cos (x) (e^{u} \frac{d}{d x} [- x]))$

Step 1.1.4.3

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

$2 (e^{- x} (- \sin (x)) + \cos (x) (e^{- x} \frac{d}{d x} [- x]))$

Step 1.1.5

Differentiate.

Tap for more steps...

Step 1.1.5.1

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

$2 (e^{- x} (- \sin (x)) + \cos (x) (e^{- x} (- \frac{d}{d x} [x])))$

Step 1.1.5.2

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

$2 (e^{- x} (- \sin (x)) + \cos (x) (e^{- x} (- 1 \cdot 1)))$

Step 1.1.5.3

Simplify the expression.

Tap for more steps...

Step 1.1.5.3.1

Multiply $- 1$ by $1$ .

$2 (e^{- x} (- \sin (x)) + \cos (x) (e^{- x} \cdot - 1))$

Step 1.1.5.3.2

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

$2 (e^{- x} (- \sin (x)) + \cos (x) (- 1 \cdot e^{- x}))$

Step 1.1.5.3.3

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

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

Step 1.1.6

Simplify.

Tap for more steps...

Step 1.1.6.1

Apply the distributive property.

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

Step 1.1.6.2

Combine terms.

Tap for more steps...

Step 1.1.6.2.1

Multiply $- 1$ by $2$ .

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

Step 1.1.6.2.2

Multiply $- 1$ by $2$ .

$- 2 e^{- x} \sin (x) - 2 \cos (x) e^{- x}$

Step 1.1.6.3

Reorder terms.

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

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- 2 e^{- x} \sin (x) - 2 e^{- x} \cos (x)$ .

$- 2 e^{- x} \sin (x) - 2 e^{- x} \cos (x)$

Step 2

Set the first derivative equal to $0$ then solve the equation $- 2 e^{- x} \sin (x) - 2 e^{- x} \cos (x) = 0$ .

Tap for more steps...

Step 2.1

Set the first derivative equal to $0$ .

$- 2 e^{- x} \sin (x) - 2 e^{- x} \cos (x) = 0$

Step 2.2

Factor $- 2 e^{- x} \sin (x) - 2 e^{- x} \cos (x)$ .

Tap for more steps...

Step 2.2.1

Factor $2 e^{- x}$ out of $- 2 e^{- x} \sin (x) - 2 e^{- x} \cos (x)$ .

Tap for more steps...

Step 2.2.1.1

Factor $2 e^{- x}$ out of $- 2 e^{- x} \sin (x)$ .

$2 e^{- x} (- 1 \sin (x)) - 2 e^{- x} \cos (x) = 0$

Step 2.2.1.2

Factor $2 e^{- x}$ out of $- 2 e^{- x} \cos (x)$ .

$2 e^{- x} (- 1 \sin (x)) + 2 e^{- x} (- 1 \cos (x)) = 0$

Step 2.2.1.3

Factor $2 e^{- x}$ out of $2 e^{- x} (- 1 \sin (x)) + 2 e^{- x} (- 1 \cos (x))$ .

$2 e^{- x} (- 1 \sin (x) - 1 \cos (x)) = 0$

Step 2.2.2

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

$2 e^{- x} (- \sin (x) - 1 \cos (x)) = 0$

Step 2.2.3

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

$2 e^{- x} (- \sin (x) - \cos (x)) = 0$

Step 2.3

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 2.4

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

Tap for more steps...

Step 2.4.1

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

$e^{- x} = 0$

Step 2.4.2

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

Tap for more steps...

Step 2.4.2.1

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

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

Step 2.4.2.2

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

Undefined

Step 2.4.2.3

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

No solution

Step 2.5

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

Tap for more steps...

Step 2.5.1

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

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

Step 2.5.2

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

Tap for more steps...

Step 2.5.2.1

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

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

Step 2.5.2.2

Separate fractions.

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

Step 2.5.2.3

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 2.5.2.4

Divide $- 1$ by $1$ .

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

Step 2.5.2.5

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

Tap for more steps...

Step 2.5.2.5.1

Cancel the common factor.

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

Step 2.5.2.5.2

Divide $- 1$ by $1$ .

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

Step 2.5.2.6

Separate fractions.

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

Step 2.5.2.7

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

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

Step 2.5.2.8

Divide $0$ by $1$ .

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

Step 2.5.2.9

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

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

Step 2.5.2.10

Add $1$ to both sides of the equation.

$- \tan (x) = 1$

Step 2.5.2.11

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

Tap for more steps...

Step 2.5.2.11.1

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

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

Step 2.5.2.11.2

Simplify the left side.

Tap for more steps...

Step 2.5.2.11.2.1

Dividing two negative values results in a positive value.

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

Step 2.5.2.11.2.2

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

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

Step 2.5.2.11.3

Simplify the right side.

Tap for more steps...

Step 2.5.2.11.3.1

Divide $1$ by $- 1$ .

$\tan (x) = - 1$

Step 2.5.2.12

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

$x = \arctan (- 1)$

Step 2.5.2.13

Simplify the right side.

Tap for more steps...

Step 2.5.2.13.1

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

$x = - \frac{π}{4}$

Step 2.5.2.14

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

$x = - \frac{π}{4} - π$

Step 2.5.2.15

Simplify the expression to find the second solution.

Tap for more steps...

Step 2.5.2.15.1

Add $2 π$ to $- \frac{π}{4} - π$ .

$x = - \frac{π}{4} - π + 2 π$

Step 2.5.2.15.2

The resulting angle of $\frac{3 π}{4}$ is positive and coterminal with $- \frac{π}{4} - π$ .

$x = \frac{3 π}{4}$

Step 2.5.2.16

Find the period of $\tan (x)$ .

Tap for more steps...

Step 2.5.2.16.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 2.5.2.16.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 2.5.2.16.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 2.5.2.16.4

Divide $π$ by $1$ .

$π$

Step 2.5.2.17

Add $π$ to every negative angle to get positive angles.

Tap for more steps...

Step 2.5.2.17.1

Add $π$ to $- \frac{π}{4}$ to find the positive angle.

$- \frac{π}{4} + π$

Step 2.5.2.17.2

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

$π \cdot \frac{4}{4} - \frac{π}{4}$

Step 2.5.2.17.3

Combine fractions.

Tap for more steps...

Step 2.5.2.17.3.1

Combine $π$ and $\frac{4}{4}$ .

$\frac{π \cdot 4}{4} - \frac{π}{4}$

Step 2.5.2.17.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 4 - π}{4}$

Step 2.5.2.17.4

Simplify the numerator.

Tap for more steps...

Step 2.5.2.17.4.1

Move $4$ to the left of $π$ .

$\frac{4 \cdot π - π}{4}$

Step 2.5.2.17.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{4}$

Step 2.5.2.17.5

List the new angles.

$x = \frac{3 π}{4}$

Step 2.5.2.18

The period of the $\tan (x)$ function is $π$ so values will repeat every $π$ radians in both directions.

$x = \frac{3 π}{4} + π n, \frac{3 π}{4} + π n$ , for any integer $n$

Step 2.6

The final solution is all the values that make $2 e^{- x} (- \sin (x) - \cos (x)) = 0$ true.

$x = \frac{3 π}{4} + π n$ , for any integer $n$

Step 3

The values which make the derivative equal to $0$ are $\frac{3 π}{4} + π n$ .

$\frac{3 π}{4} + π n$

Step 4

After finding the point that makes the derivative $f' (x) = - 2 e^{- x} \sin (x) - 2 e^{- x} \cos (x)$ equal to $0$ or undefined, the interval to check where $f (x) = 2 e^{- x} \cos (x)$ is increasing and where it is decreasing is $(- \infty, \frac{3 π}{4} + π n) \cup (\frac{3 π}{4} + π n, \infty)$ .

$(- \infty, \frac{3 π}{4} + π n) \cup (\frac{3 π}{4} + π n, \infty)$

Step 5

Substitute a value from the interval $(- \infty, \frac{3 π}{4} + π n)$ into the derivative to determine if the function is increasing or decreasing.

Tap for more steps...

Step 5.1

Replace the variable $x$ with $\frac{3 π}{4} + π n - 1$ in the expression.

$f' (\frac{3 π}{4} + π n - 1) = - 2 e^{- (\frac{3 π}{4} + π n - 1)} \sin (\frac{3 π}{4} + π n - 1) - 2 e^{- (\frac{3 π}{4} + π n - 1)} \cos (\frac{3 π}{4} + π n - 1)$

Step 5.2

Simplify the result.

Tap for more steps...

Step 5.2.1

Simplify each term.

Tap for more steps...

Step 5.2.1.1

Apply the distributive property.

$f' (\frac{3 π}{4} + π n - 1) = - 2 e^{- \frac{3 π}{4} - (π n) + 1} \sin (\frac{3 π}{4} + π n - 1) - 2 e^{- (\frac{3 π}{4} + π n - 1)} \cos (\frac{3 π}{4} + π n - 1)$

Step 5.2.1.2

Multiply $- 1$ by $- 1$ .

$f' (\frac{3 π}{4} + π n - 1) = - 2 e^{- \frac{3 π}{4} - π n + 1} \sin (\frac{3 π}{4} + π n - 1) - 2 e^{- (\frac{3 π}{4} + π n - 1)} \cos (\frac{3 π}{4} + π n - 1)$

Step 5.2.1.3

Apply the distributive property.

$f' (\frac{3 π}{4} + π n - 1) = - 2 e^{- \frac{3 π}{4} - π n + 1} \sin (\frac{3 π}{4} + π n - 1) - 2 e^{- \frac{3 π}{4} - (π n) + 1} \cos (\frac{3 π}{4} + π n - 1)$

Step 5.2.1.4

Multiply $- 1$ by $- 1$ .

$f' (\frac{3 π}{4} + π n - 1) = - 2 e^{- \frac{3 π}{4} - π n + 1} \sin (\frac{3 π}{4} + π n - 1) - 2 e^{- \frac{3 π}{4} - π n + 1} \cos (\frac{3 π}{4} + π n - 1)$

Step 5.2.2

The final answer is $- 2 e^{- \frac{3 π}{4} - π n + 1} \sin (\frac{3 π}{4} + π n - 1) - 2 e^{- \frac{3 π}{4} - π n + 1} \cos (\frac{3 π}{4} + π n - 1)$ .

$- 2 e^{- \frac{3 π}{4} - π n + 1} \sin (\frac{3 π}{4} + π n - 1) - 2 e^{- \frac{3 π}{4} - π n + 1} \cos (\frac{3 π}{4} + π n - 1)$

Step 5.3

At $x = \frac{3 π}{4} + π n - 1$ the derivative is $- 2 e^{- \frac{3 π}{4} - π n + 1} \sin (\frac{3 π}{4} + π n - 1) - 2 e^{- \frac{3 π}{4} - π n + 1} \cos (\frac{3 π}{4} + π n - 1)$ . Since this is negative, the function is decreasing on $(- \infty, \frac{3 π}{4} + π n)$ .

Decreasing on $(- \infty, \frac{3 π}{4} + π n)$ since $f' (x) < 0$

Step 6

Substitute a value from the interval $(\frac{3 π}{4} + π n, \infty)$ into the derivative to determine if the function is increasing or decreasing.

Tap for more steps...

Step 6.1

Replace the variable $x$ with $\frac{3 π}{4} + π n + 1$ in the expression.

$f' (\frac{3 π}{4} + π n + 1) = - 2 e^{- (\frac{3 π}{4} + π n + 1)} \sin (\frac{3 π}{4} + π n + 1) - 2 e^{- (\frac{3 π}{4} + π n + 1)} \cos (\frac{3 π}{4} + π n + 1)$

Step 6.2

Simplify the result.

Tap for more steps...

Step 6.2.1

Simplify each term.

Tap for more steps...

Step 6.2.1.1

Apply the distributive property.

$f' (\frac{3 π}{4} + π n + 1) = - 2 e^{- \frac{3 π}{4} - (π n) - 1 \cdot 1} \sin (\frac{3 π}{4} + π n + 1) - 2 e^{- (\frac{3 π}{4} + π n + 1)} \cos (\frac{3 π}{4} + π n + 1)$

Step 6.2.1.2

Multiply $- 1$ by $1$ .

$f' (\frac{3 π}{4} + π n + 1) = - 2 e^{- \frac{3 π}{4} - π n - 1} \sin (\frac{3 π}{4} + π n + 1) - 2 e^{- (\frac{3 π}{4} + π n + 1)} \cos (\frac{3 π}{4} + π n + 1)$

Step 6.2.1.3

Apply the distributive property.

$f' (\frac{3 π}{4} + π n + 1) = - 2 e^{- \frac{3 π}{4} - π n - 1} \sin (\frac{3 π}{4} + π n + 1) - 2 e^{- \frac{3 π}{4} - (π n) - 1 \cdot 1} \cos (\frac{3 π}{4} + π n + 1)$

Step 6.2.1.4

Multiply $- 1$ by $1$ .

$f' (\frac{3 π}{4} + π n + 1) = - 2 e^{- \frac{3 π}{4} - π n - 1} \sin (\frac{3 π}{4} + π n + 1) - 2 e^{- \frac{3 π}{4} - π n - 1} \cos (\frac{3 π}{4} + π n + 1)$

Step 6.2.2

The final answer is $- 2 e^{- \frac{3 π}{4} - π n - 1} \sin (\frac{3 π}{4} + π n + 1) - 2 e^{- \frac{3 π}{4} - π n - 1} \cos (\frac{3 π}{4} + π n + 1)$ .

$- 2 e^{- \frac{3 π}{4} - π n - 1} \sin (\frac{3 π}{4} + π n + 1) - 2 e^{- \frac{3 π}{4} - π n - 1} \cos (\frac{3 π}{4} + π n + 1)$

Step 6.3

At $x = \frac{3 π}{4} + π n + 1$ the derivative is $- 2 e^{- \frac{3 π}{4} - π n - 1} \sin (\frac{3 π}{4} + π n + 1) - 2 e^{- \frac{3 π}{4} - π n - 1} \cos (\frac{3 π}{4} + π n + 1)$ . Since this is negative, the function is decreasing on $(\frac{3 π}{4} + π n, \infty)$ .

Decreasing on $(\frac{3 π}{4} + π n, \infty)$ since $f' (x) < 0$

Step 7

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

Decreasing on: $(- \infty, \frac{3 π}{4} + π n), (\frac{3 π}{4} + π n, \infty)$

Step 8