Calculus Examples

$f (x) = x^{4} e^{- 2 x}$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

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 $f (x) = x^{4}$ and $g (x) = e^{- 2 x}$ .

$x^{4} \frac{d}{d x} [e^{- 2 x}] + e^{- 2 x} \frac{d}{d x} [x^{4}]$

Step 1.1.2

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 $g (x) = - 2 x$ .

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Step 1.1.2.1

To apply the Chain Rule, set $u$ as $- 2 x$ .

$x^{4} (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- 2 x]) + e^{- 2 x} \frac{d}{d x} [x^{4}]$

Step 1.1.2.2

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

$x^{4} (e^{u} \frac{d}{d x} [- 2 x]) + e^{- 2 x} \frac{d}{d x} [x^{4}]$

Step 1.1.2.3

Replace all occurrences of $u$ with $- 2 x$ .

$x^{4} (e^{- 2 x} \frac{d}{d x} [- 2 x]) + e^{- 2 x} \frac{d}{d x} [x^{4}]$

Step 1.1.3

Differentiate.

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Step 1.1.3.1

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

$x^{4} (e^{- 2 x} (- 2 \frac{d}{d x} [x])) + e^{- 2 x} \frac{d}{d x} [x^{4}]$

Step 1.1.3.2

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

$x^{4} (e^{- 2 x} (- 2 \cdot 1)) + e^{- 2 x} \frac{d}{d x} [x^{4}]$

Step 1.1.3.3

Simplify the expression.

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Step 1.1.3.3.1

Multiply $- 2$ by $1$ .

$x^{4} (e^{- 2 x} \cdot - 2) + e^{- 2 x} \frac{d}{d x} [x^{4}]$

Step 1.1.3.3.2

Move $- 2$ to the left of $e^{- 2 x}$ .

$x^{4} (- 2 \cdot e^{- 2 x}) + e^{- 2 x} \frac{d}{d x} [x^{4}]$

Step 1.1.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$x^{4} (- 2 e^{- 2 x}) + e^{- 2 x} (4 x^{3})$

Step 1.1.4

Simplify.

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Step 1.1.4.1

Reorder terms.

$- 2 e^{- 2 x} x^{4} + 4 e^{- 2 x} x^{3}$

Step 1.1.4.2

Reorder factors in $- 2 e^{- 2 x} x^{4} + 4 e^{- 2 x} x^{3}$ .

$f' (x) = - 2 x^{4} e^{- 2 x} + 4 x^{3} e^{- 2 x}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $- 2 x^{4} e^{- 2 x} + 4 x^{3} e^{- 2 x}$ .

$- 2 x^{4} e^{- 2 x} + 4 x^{3} e^{- 2 x}$

Step 2

Set the first derivative equal to $0$ then solve the equation $- 2 x^{4} e^{- 2 x} + 4 x^{3} e^{- 2 x} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$- 2 x^{4} e^{- 2 x} + 4 x^{3} e^{- 2 x} = 0$

Step 2.2

Factor $2 x^{3} e^{- 2 x}$ out of $- 2 x^{4} e^{- 2 x} + 4 x^{3} e^{- 2 x}$ .

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Step 2.2.1

Factor $2 x^{3} e^{- 2 x}$ out of $- 2 x^{4} e^{- 2 x}$ .

$2 x^{3} e^{- 2 x} (- x) + 4 x^{3} e^{- 2 x} = 0$

Step 2.2.2

Factor $2 x^{3} e^{- 2 x}$ out of $4 x^{3} e^{- 2 x}$ .

$2 x^{3} e^{- 2 x} (- x) + 2 x^{3} e^{- 2 x} (2) = 0$

Step 2.2.3

Factor $2 x^{3} e^{- 2 x}$ out of $2 x^{3} e^{- 2 x} (- x) + 2 x^{3} e^{- 2 x} (2)$ .

$2 x^{3} e^{- 2 x} (- x + 2) = 0$

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 $0$ .

$x^{3} = 0$

$e^{- 2 x} = 0$

$- x + 2 = 0$

Step 2.4

Set $x^{3}$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x^{3}$ equal to $0$ .

$x^{3} = 0$

Step 2.4.2

Solve $x^{3} = 0$ for $x$ .

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Step 2.4.2.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{0}$

Step 2.4.2.2

Simplify $\sqrt[3]{0}$ .

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Step 2.4.2.2.1

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Step 2.4.2.2.2

Pull terms out from under the radical, assuming real numbers.

$x = 0$

Step 2.5

Set $e^{- 2 x}$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $e^{- 2 x}$ equal to $0$ .

$e^{- 2 x} = 0$

Step 2.5.2

Solve $e^{- 2 x} = 0$ for $x$ .

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Step 2.5.2.1

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

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

Step 2.5.2.2

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

Undefined

Step 2.5.2.3

There is no solution for $e^{- 2 x} = 0$

No solution

Step 2.6

Set $- x + 2$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $- x + 2$ equal to $0$ .

$- x + 2 = 0$

Step 2.6.2

Solve $- x + 2 = 0$ for $x$ .

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Step 2.6.2.1

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 2.6.2.2

Divide each term in $- x = - 2$ by $- 1$ and simplify.

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Step 2.6.2.2.1

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Step 2.6.2.2.2

Simplify the left side.

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Step 2.6.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Step 2.6.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Step 2.6.2.2.3

Simplify the right side.

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Step 2.6.2.2.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 2.7

The final solution is all the values that make $2 x^{3} e^{- 2 x} (- x + 2) = 0$ true.

$x = 0, 2$

Step 3

Find the values where the derivative is undefined.

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Step 3.1

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 4

Evaluate $x^{4} e^{- 2 x}$ at each $x$ value where the derivative is $0$ or undefined.

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Step 4.1

Evaluate at $x = 0$ .

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Step 4.1.1

Substitute $0$ for $x$ .

${(0)}^{4} e^{- 2 \cdot 0}$

Step 4.1.2

Simplify.

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Step 4.1.2.1

Raising $0$ to any positive power yields $0$ .

$0 e^{- 2 \cdot 0}$

Step 4.1.2.2

Multiply $- 2$ by $0$ .

$0 e^{0}$

Step 4.1.2.3

Anything raised to $0$ is $1$ .

$0 \cdot 1$

Step 4.1.2.4

Multiply $0$ by $1$ .

$0$

Step 4.2

Evaluate at $x = 2$ .

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Step 4.2.1

Substitute $2$ for $x$ .

${(2)}^{4} e^{- 2 \cdot 2}$

Step 4.2.2

Simplify.

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Step 4.2.2.1

Raise $2$ to the power of $4$ .

$16 e^{- 2 \cdot 2}$

Step 4.2.2.2

Multiply $- 2$ by $2$ .

$16 e^{- 4}$

Step 4.2.2.3

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

$16 \frac{1}{e^{4}}$

Step 4.2.2.4

Combine $16$ and $\frac{1}{e^{4}}$ .

$\frac{16}{e^{4}}$

Step 4.3

List all of the points.

$(0, 0), (2, \frac{16}{e^{4}})$

Step 5