Calculus Examples

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

Step 1

Find the derivative.

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

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

Step 1.2

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 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.3

Differentiate.

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Simplify the expression.

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Step 1.3.3.1

Multiply $- 1$ by $1$ .

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

Step 1.3.3.2

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

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

Step 1.3.3.3

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

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

Step 1.3.4

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 1.3.5

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

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

Step 1.4

Simplify.

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Step 1.4.1

Reorder terms.

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

Step 1.4.2

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

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

Step 2

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

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Step 2.1

Factor $e^{- x}$ out of $- x e^{- x} + e^{- x}$ .

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Step 2.1.1

Factor $e^{- x}$ out of $- x e^{- x}$ .

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

Step 2.1.2

Multiply by $1$ .

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

Step 2.1.3

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

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

Step 2.2

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 2.3

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

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Step 2.3.1

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

$e^{- x} = 0$

Step 2.3.2

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

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Step 2.3.2.1

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

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

Step 2.3.2.2

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

Undefined

Step 2.3.2.3

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

No solution

Step 2.4

Set $- x + 1$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $- x + 1$ equal to $0$ .

$- x + 1 = 0$

Step 2.4.2

Solve $- x + 1 = 0$ for $x$ .

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Step 2.4.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 2.4.2.2

Divide each term in $- x = - 1$ by $- 1$ and simplify.

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Step 2.4.2.2.1

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Step 2.4.2.2.2

Simplify the left side.

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Step 2.4.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Step 2.4.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Step 2.4.2.2.3

Simplify the right side.

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Step 2.4.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 2.5

The final solution is all the values that make $e^{- x} (- x + 1) = 0$ true.

$x = 1$

Step 3

Solve the original function $f (x) = x e^{- x}$ at $x = 1$ .

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Step 3.1

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

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

Step 3.2

Simplify the result.

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Step 3.2.1

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

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

Step 3.2.2

Multiply $- 1$ by $1$ .

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

Step 3.2.3

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

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

Step 3.2.4

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

$\frac{1}{e}$

Step 4

The horizontal tangent line on function $f (x) = x e^{- x}$ is $y = \frac{1}{e}$ .

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

Step 5