Calculus Examples

$(1 - x) e^{x}$

Step 1

Write $(1 - x) e^{x}$ as a function.

$f (x) = (1 - x) e^{x}$

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 $f (x) = 1 - x$ and $g (x) = e^{x}$ .

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

Step 2.1.2

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

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

Step 2.1.3

Differentiate.

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Step 2.1.3.1

By the Sum Rule, the derivative of $1 - x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- x]$ .

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

Step 2.1.3.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$(1 - x) e^{x} + e^{x} (0 + \frac{d}{d x} [- x])$

Step 2.1.3.3

Add $0$ and $\frac{d}{d x} [- x]$ .

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

Step 2.1.3.4

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

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

Step 2.1.3.5

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

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

Step 2.1.3.6

Simplify the expression.

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Step 2.1.3.6.1

Multiply $- 1$ by $1$ .

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

Step 2.1.3.6.2

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

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

Step 2.1.3.6.3

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

$(1 - x) e^{x} - e^{x}$

Step 2.1.4

Simplify.

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Step 2.1.4.1

Apply the distributive property.

$1 e^{x} - x e^{x} - e^{x}$

Step 2.1.4.2

Combine terms.

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Step 2.1.4.2.1

Multiply $e^{x}$ by $1$ .

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

Step 2.1.4.2.2

Subtract $e^{x}$ from $e^{x}$ .

$- x e^{x} + 0$

Step 2.1.4.2.3

Add $- x e^{x}$ and $0$ .

$- x e^{x}$

Step 2.1.4.3

Reorder the factors of $- x e^{x}$ .

$- e^{x} x$

Step 2.1.4.4

Reorder factors in $- e^{x} x$ .

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

Step 2.2

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

$- x e^{x}$

Step 3

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

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Step 3.1

Set the first derivative equal to $0$ .

$- x e^{x} = 0$

Step 3.2

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

$x = 0$

$e^{x} = 0$

Step 3.3

Set $x$ equal to $0$ .

$x = 0$

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

The final solution is all the values that make $- x e^{x} = 0$ true.

$x = 0$

Step 4

The values which make the derivative equal to $0$ are $0$ .

$0$

Step 5

After finding the point that makes the derivative $f' (x) = - x e^{x}$ equal to $0$ or undefined, the interval to check where $f (x) = (1 - x) e^{x}$ is increasing and where it is decreasing is $(- \infty, 0) \cup (0, \infty)$ .

$(- \infty, 0) \cup (0, \infty)$

Step 6

Substitute a value from the interval $(- \infty, 0)$ into the derivative to determine if the function is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $- 1$ in the expression.

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

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $e^{- 1}$ by $1$ .

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

Step 6.2.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 6.2.3

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

$\frac{1}{e}$

Step 6.3

At $x = - 1$ the derivative is $\frac{1}{e}$ . Since this is positive, the function is increasing on $(- \infty, 0)$ .

Increasing on $(- \infty, 0)$ since $f' (x) > 0$

Step 7

Substitute a value from the interval $(0, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $1$ in the expression.

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

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify.

$f' (1) = - 1 \cdot e$

Step 7.2.2

Rewrite $- 1 e$ as $- e$ .

$f' (1) = - e$

Step 7.2.3

The final answer is $- e$ .

$- e$

Step 7.3

At $x = 1$ the derivative is $- e$ . Since this is negative, the function is decreasing on $(0, \infty)$ .

Decreasing on $(0, \infty)$ since $f' (x) < 0$

Step 8

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

Increasing on: $(- \infty, 0)$

Decreasing on: $(0, \infty)$

Step 9