Calculus Examples

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

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 $f (x) = x$ 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

Find the second derivative of the function.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

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

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

Step 2.2.3

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 2.2.3.1

To apply the Chain Rule, set $u_{1}$ as $- x$ .

$f'' (x) = - (x (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x)) + \frac{d}{d x} (e^{- x})$

Step 2.2.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$f'' (x) = - (x (e^{u_{1}} \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x)) + \frac{d}{d x} (e^{- x})$

Step 2.2.3.3

Replace all occurrences of $u_{1}$ with $- x$ .

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

Multiply $- 1$ by $1$ .

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

Step 2.2.8

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

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

Step 2.2.9

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

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

Step 2.2.10

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

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

Step 2.3

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

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Step 2.3.1

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 2.3.1.1

To apply the Chain Rule, set $u_{2}$ as $- x$ .

$f'' (x) = - (x (- e^{- x}) + e^{- x}) + \frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- x)$

Step 2.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

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

Step 2.3.1.3

Replace all occurrences of $u_{2}$ with $- x$ .

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Multiply $- 1$ by $1$ .

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

Step 2.3.5

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

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

Step 2.3.6

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

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

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

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

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Multiply $- 1$ by $- 1$ .

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

Step 2.4.2.2

Multiply $x$ by $1$ .

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

Step 2.4.2.3

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

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

Step 2.4.3

Reorder terms.

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

Step 2.4.4

Reorder factors in $e^{- x} x - 2 e^{- x}$ .

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

Step 3

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

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

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

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

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

Step 4.1.2

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 4.1.2.1

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

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

Step 4.1.2.2

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

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

Step 4.1.2.3

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

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

Step 4.1.3

Differentiate.

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Step 4.1.3.1

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

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

Step 4.1.3.2

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

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

Step 4.1.3.3

Simplify the expression.

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Step 4.1.3.3.1

Multiply $- 1$ by $1$ .

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

Step 4.1.3.3.2

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

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

Step 4.1.3.3.3

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

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

Step 4.1.3.4

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

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

Step 4.1.3.5

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

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

Step 4.1.4

Simplify.

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Step 4.1.4.1

Reorder terms.

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

Step 4.1.4.2

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

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

Step 4.2

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

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

Step 5

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

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Step 5.1

Set the first derivative equal to $0$ .

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

Multiply by $1$ .

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

Step 5.2.3

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

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

Step 5.3

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 5.4

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

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Step 5.4.1

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

$e^{- x} = 0$

Step 5.4.2

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

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Step 5.4.2.1

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

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

Step 5.4.2.2

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

Undefined

Step 5.4.2.3

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

No solution

Step 5.5

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

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Step 5.5.1

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

$- x + 1 = 0$

Step 5.5.2

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

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Step 5.5.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 5.5.2.2

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

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Step 5.5.2.2.1

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

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

Step 5.5.2.2.2

Simplify the left side.

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Step 5.5.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 5.5.2.2.2.2

Divide $x$ by $1$ .

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

Step 5.5.2.2.3

Simplify the right side.

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Step 5.5.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 5.6

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

$x = 1$

Step 6

Find the values where the derivative is undefined.

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Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = 1$

Step 8

Evaluate the second derivative at $x = 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$(1) e^{- (1)} - 2 e^{- (1)}$

Step 9

Evaluate the second derivative.

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Step 9.1

Simplify each term.

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Step 9.1.1

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

$e^{- (1)} - 2 e^{- (1)}$

Step 9.1.2

Multiply $- 1$ by $1$ .

$e^{- 1} - 2 e^{- (1)}$

Step 9.1.3

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

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

Step 9.1.4

Multiply $- 1$ by $1$ .

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

Step 9.1.5

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

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

Step 9.1.6

Combine $- 2$ and $\frac{1}{e}$ .

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

Step 9.1.7

Move the negative in front of the fraction.

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

Step 9.2

Combine fractions.

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Step 9.2.1

Combine the numerators over the common denominator.

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

Step 9.2.2

Simplify the expression.

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Step 9.2.2.1

Subtract $2$ from $1$ .

$\frac{- 1}{e}$

Step 9.2.2.2

Move the negative in front of the fraction.

$- \frac{1}{e}$

Step 10

$x = 1$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 1$ is a local maximum

Step 11

Find the y-value when $x = 1$ .

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Step 11.1

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

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

Step 11.2

Simplify the result.

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Step 11.2.1

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

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

Step 11.2.2

Multiply $- 1$ by $1$ .

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

Step 11.2.3

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

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

Step 11.2.4

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

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

Step 12

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

$(1, \frac{1}{e})$ is a local maxima

Step 13