Calculus Examples

$x e^{- 2 x}$

Step 1

Find the first derivative.

Tap for more steps...

Step 1.1

Find the first derivative.

Tap for more steps...

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

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

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 $g (x) = - 2 x$ .

Tap for more steps...

Step 1.1.2.1

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

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

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 $a$ = $e$ .

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

Step 1.1.2.3

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

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

Step 1.1.3

Differentiate.

Tap for more steps...

Step 1.1.3.1

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

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

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 $n = 1$ .

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

Step 1.1.3.3

Simplify the expression.

Tap for more steps...

Step 1.1.3.3.1

Multiply $- 2$ by $1$ .

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

Step 1.1.3.3.2

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

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

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 $n = 1$ .

$x (- 2 e^{- 2 x}) + e^{- 2 x} \cdot 1$

Step 1.1.3.5

Multiply $e^{- 2 x}$ by $1$ .

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

Step 1.1.4

Simplify.

Tap for more steps...

Step 1.1.4.1

Reorder terms.

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

Step 1.1.4.2

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

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

Step 1.2

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

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

Step 2

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

Tap for more steps...

Step 2.1

Set the first derivative equal to $0$ .

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

Step 2.2

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

Tap for more steps...

Step 2.2.1

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

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

Step 2.2.2

Multiply by $1$ .

$e^{- 2 x} (- 2 x) + e^{- 2 x} \cdot 1 = 0$

Step 2.2.3

Factor $e^{- 2 x}$ out of $e^{- 2 x} (- 2 x) + e^{- 2 x} \cdot 1$ .

$e^{- 2 x} (- 2 x + 1) = 0$

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 $0$ .

$e^{- 2 x} = 0$

$- 2 x + 1 = 0$

Step 2.4

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

Tap for more steps...

Step 2.4.1

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

$e^{- 2 x} = 0$

Step 2.4.2

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

Tap for more steps...

Step 2.4.2.1

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

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

Step 2.4.2.2

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

Undefined

Step 2.4.2.3

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

No solution

Step 2.5

Set $- 2 x + 1$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.5.1

Set $- 2 x + 1$ equal to $0$ .

$- 2 x + 1 = 0$

Step 2.5.2

Solve $- 2 x + 1 = 0$ for $x$ .

Tap for more steps...

Step 2.5.2.1

Subtract $1$ from both sides of the equation.

$- 2 x = - 1$

Step 2.5.2.2

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

Tap for more steps...

Step 2.5.2.2.1

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

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

Step 2.5.2.2.2

Simplify the left side.

Tap for more steps...

Step 2.5.2.2.2.1

Cancel the common factor of $- 2$ .

Tap for more steps...

Step 2.5.2.2.2.1.1

Cancel the common factor.

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

Step 2.5.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.5.2.2.3

Simplify the right side.

Tap for more steps...

Step 2.5.2.2.3.1

Dividing two negative values results in a positive value.

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

Step 2.6

The final solution is all the values that make $e^{- 2 x} (- 2 x + 1) = 0$ true.

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

Step 3

Find the values where the derivative is undefined.

Tap for more steps...

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

Tap for more steps...

Step 4.1

Evaluate at $x = \frac{1}{2}$ .

Tap for more steps...

Step 4.1.1

Substitute $\frac{1}{2}$ for $x$ .

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

Step 4.1.2

Simplify.

Tap for more steps...

Step 4.1.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 4.1.2.1.1

Factor $2$ out of $- 2$ .

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

Step 4.1.2.1.2

Cancel the common factor.

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

Step 4.1.2.1.3

Rewrite the expression.

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

Step 4.1.2.2

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

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

Step 4.1.2.3

Multiply $\frac{1}{2}$ by $\frac{1}{e}$ .

$\frac{1}{2 e}$

Step 4.2

List all of the points.

$(\frac{1}{2}, \frac{1}{2 e})$

Step 5