Calculus Examples

$x e^{x}$

Step 1

Write

$x e^{x}$ as a function.

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

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

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

$f (x) = x$ and

$g (x) = e^{x}$ .

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

Step 2.1.2

Differentiate using the Exponential Rule which states that

$\frac{d}{d x} [a^{x}]$ is

$a^{x} \ln (a)$ where

$a$ =

$e$ .

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

Step 2.1.3

Differentiate using the Power Rule.

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Step 2.1.3.1

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 2.1.3.2

Multiply

$e^{x}$ by

$1$ .

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

Step 2.2

The first derivative of

$f (x)$ with respect to

$x$ is

$x e^{x} + e^{x}$ .

$x e^{x} + e^{x}$

Step 3

Set the first derivative equal to

$0$ then solve the equation

$x e^{x} + e^{x} = 0$ .

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Step 3.1

Set the first derivative equal to

$0$ .

$x e^{x} + e^{x} = 0$

Step 3.2

Factor

$e^{x}$ out of

$x e^{x} + e^{x}$ .

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Step 3.2.1

Factor

$e^{x}$ out of

$x e^{x}$ .

$e^{x} x + e^{x} = 0$

Step 3.2.2

Multiply by

$1$ .

$e^{x} x + e^{x} \cdot 1 = 0$

Step 3.2.3

Factor

$e^{x}$ out of

$e^{x} x + e^{x} \cdot 1$ .

$e^{x} (x + 1) = 0$

Step 3.3

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$e^{x} = 0$

$x + 1 = 0$

Step 3.4

Set

$e^{x}$ equal to

$0$ and solve for

$x$ .

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Step 3.4.1

Set

$e^{x}$ equal to

$0$ .

$e^{x} = 0$

Step 3.4.2

Solve

$e^{x} = 0$ for

$x$ .

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Step 3.4.2.1

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

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

Step 3.4.2.2

The equation cannot be solved because

$\ln (0)$ is undefined.

Undefined

Step 3.4.2.3

There is no solution for

$e^{x} = 0$

No solution

Step 3.5

Set

$x + 1$ equal to

$0$ and solve for

$x$ .

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Step 3.5.1

Set

$x + 1$ equal to

$0$ .

$x + 1 = 0$

Step 3.5.2

Subtract

$1$ from both sides of the equation.

$x = - 1$

Step 3.6

The final solution is all the values that make

$e^{x} (x + 1) = 0$ true.

$x = - 1$

Step 4

The values which make the derivative equal to

$0$ are

$- 1$ .

$- 1$

Step 5

After finding the point that makes the derivative

$f' (x) = x e^{x} + e^{x}$ equal to

$0$ or undefined, the interval to check where

$f (x) = x e^{x}$ is increasing and where it is decreasing is

$(- \infty, - 1) \cup (- 1, \infty)$ .

$(- \infty, - 1) \cup (- 1, \infty)$

Step 6

Substitute a value from the interval

$(- \infty, - 1)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable

$x$ with

$- 2$ in the expression.

$f' (- 2) = (- 2) \cdot e^{- 2} + e^{- 2}$

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

Rewrite the expression using the negative exponent rule

$b^{- n} = \frac{1}{b^{n}}$ .

$f' (- 2) = - 2 \cdot \frac{1}{e^{2}} + e^{- 2}$

Step 6.2.1.2

Combine

$- 2$ and

$\frac{1}{e^{2}}$ .

$f' (- 2) = \frac{- 2}{e^{2}} + e^{- 2}$

Step 6.2.1.3

Move the negative in front of the fraction.

$f' (- 2) = - \frac{2}{e^{2}} + e^{- 2}$

Step 6.2.1.4

Rewrite the expression using the negative exponent rule

$b^{- n} = \frac{1}{b^{n}}$ .

$f' (- 2) = - \frac{2}{e^{2}} + \frac{1}{e^{2}}$

Step 6.2.2

Combine fractions.

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Step 6.2.2.1

Combine the numerators over the common denominator.

$f' (- 2) = \frac{- 2 + 1}{e^{2}}$

Step 6.2.2.2

Simplify the expression.

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Step 6.2.2.2.1

Add

$- 2$ and

$1$ .

$f' (- 2) = \frac{- 1}{e^{2}}$

Step 6.2.2.2.2

Move the negative in front of the fraction.

$f' (- 2) = - \frac{1}{e^{2}}$

Step 6.2.3

The final answer is

$- \frac{1}{e^{2}}$ .

$- \frac{1}{e^{2}}$

Step 6.3

$x = - 2$ the derivative is

$- \frac{1}{e^{2}}$ . Since this is negative, the function is decreasing on

$(- \infty, - 1)$ .

Decreasing on

$(- \infty, - 1)$ since

$f' (x) < 0$

Decreasing on

$(- \infty, - 1)$ since

$f' (x) < 0$

Step 7

Substitute a value from the interval

$(- 1, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable

$x$ with

$0$ in the expression.

$f' (0) = (0) \cdot e^{0} + e^{0}$

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

Anything raised to

$0$ is

$1$ .

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

Step 7.2.1.2

Multiply

$0$ by

$1$ .

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

Step 7.2.1.3

Anything raised to

$0$ is

$1$ .

$f' (0) = 0 + 1$

Step 7.2.2

Add

$0$ and

$1$ .

$f' (0) = 1$

Step 7.2.3

The final answer is

$1$ .

$1$

Step 7.3

$x = 0$ the derivative is

$1$ . Since this is positive, the function is increasing on

$(- 1, \infty)$ .

Increasing on

$(- 1, \infty)$ since

$f' (x) > 0$

Increasing on

$(- 1, \infty)$ since

$f' (x) > 0$

Step 8

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

Increasing on:

$(- 1, \infty)$

Decreasing on:

$(- \infty, - 1)$

Step 9