Calculus Examples

$y = x^{2} e^{x}$

Step 1

Write $y = x^{2} e^{x}$ as a function.

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

Step 2

Find the first derivative of the function.

Tap for more steps...

Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 2.4

Simplify.

Tap for more steps...

Step 2.4.1

Reorder terms.

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

Step 2.4.2

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

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

Step 3

Find the second derivative of the function.

Tap for more steps...

Step 3.1

By the Sum Rule, the derivative of $x^{2} e^{x} + 2 x e^{x}$ with respect to $x$ is $\frac{d}{d x} [x^{2} e^{x}] + \frac{d}{d x} [2 x e^{x}]$ .

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

Step 3.2

Evaluate $\frac{d}{d x} [x^{2} e^{x}]$ .

Tap for more steps...

Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 3.3

Evaluate $\frac{d}{d x} [2 x e^{x}]$ .

Tap for more steps...

Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

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

$f'' (x) = x^{2} e^{x} + e^{x} (2 x) + 2 (x e^{x} + e^{x} \cdot 1)$

Step 3.3.5

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

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

Step 3.4

Simplify.

Tap for more steps...

Step 3.4.1

Apply the distributive property.

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

Step 3.4.2

Add $e^{x} (2 x)$ and $2 x e^{x}$ .

Tap for more steps...

Step 3.4.2.1

Move $e^{x}$ .

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

Step 3.4.2.2

Add $2 x e^{x}$ and $2 x e^{x}$ .

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

Step 3.4.3

Reorder terms.

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

Step 3.4.4

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

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

Step 4

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

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

Step 5

Find the first derivative.

Tap for more steps...

Step 5.1

Find the first derivative.

Tap for more steps...

Step 5.1.1

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

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

Step 5.1.2

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

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

Step 5.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

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

Step 5.1.4

Simplify.

Tap for more steps...

Step 5.1.4.1

Reorder terms.

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

Step 5.1.4.2

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

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

Step 5.2

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

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

Step 6

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

Tap for more steps...

Step 6.1

Set the first derivative equal to $0$ .

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

Step 6.2

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

Tap for more steps...

Step 6.2.1

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

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

Step 6.2.2

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

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

Step 6.2.3

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

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

Step 6.3

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

$x = 0$

$e^{x} = 0$

$x + 2 = 0$

Step 6.4

Set $x$ equal to $0$ .

$x = 0$

Step 6.5

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

Tap for more steps...

Step 6.5.1

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

$e^{x} = 0$

Step 6.5.2

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

Tap for more steps...

Step 6.5.2.1

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

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

Step 6.5.2.2

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

Undefined

Step 6.5.2.3

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

No solution

Step 6.6

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

Tap for more steps...

Step 6.6.1

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

$x + 2 = 0$

Step 6.6.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 6.7

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

$x = 0, - 2$

Step 7

Find the values where the derivative is undefined.

Tap for more steps...

Step 7.1

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 8

Critical points to evaluate.

$x = 0, - 2$

Step 9

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

${(0)}^{2} e^{0} + 4 (0) e^{0} + 2 e^{0}$

Step 10

Evaluate the second derivative.

Tap for more steps...

Step 10.1

Simplify each term.

Tap for more steps...

Step 10.1.1

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

$0 e^{0} + 4 (0) e^{0} + 2 e^{0}$

Step 10.1.2

Anything raised to $0$ is $1$ .

$0 \cdot 1 + 4 (0) e^{0} + 2 e^{0}$

Step 10.1.3

Multiply $0$ by $1$ .

$0 + 4 (0) e^{0} + 2 e^{0}$

Step 10.1.4

Multiply $4$ by $0$ .

$0 + 0 e^{0} + 2 e^{0}$

Step 10.1.5

Anything raised to $0$ is $1$ .

$0 + 0 \cdot 1 + 2 e^{0}$

Step 10.1.6

Multiply $0$ by $1$ .

$0 + 0 + 2 e^{0}$

Step 10.1.7

Anything raised to $0$ is $1$ .

$0 + 0 + 2 \cdot 1$

Step 10.1.8

Multiply $2$ by $1$ .

$0 + 0 + 2$

Step 10.2

Simplify by adding numbers.

Tap for more steps...

Step 10.2.1

Add $0$ and $0$ .

$0 + 2$

Step 10.2.2

Add $0$ and $2$ .

$2$

Step 11

$x = 0$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 0$ is a local minimum

Step 12

Find the y-value when $x = 0$ .

Tap for more steps...

Step 12.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = {(0)}^{2} e^{0}$

Step 12.2

Simplify the result.

Tap for more steps...

Step 12.2.1

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

$f (0) = 0 e^{0}$

Step 12.2.2

Anything raised to $0$ is $1$ .

$f (0) = 0 \cdot 1$

Step 12.2.3

Multiply $0$ by $1$ .

$f (0) = 0$

Step 12.2.4

The final answer is $0$ .

$y = 0$

Step 13

Evaluate the second derivative at $x = - 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

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

Step 14

Evaluate the second derivative.

Tap for more steps...

Step 14.1

Simplify each term.

Tap for more steps...

Step 14.1.1

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

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

Step 14.1.2

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

$4 \frac{1}{e^{2}} + 4 (- 2) e^{- 2} + 2 e^{- 2}$

Step 14.1.3

Combine $4$ and $\frac{1}{e^{2}}$ .

$\frac{4}{e^{2}} + 4 (- 2) e^{- 2} + 2 e^{- 2}$

Step 14.1.4

Multiply $4$ by $- 2$ .

$\frac{4}{e^{2}} - 8 e^{- 2} + 2 e^{- 2}$

Step 14.1.5

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

$\frac{4}{e^{2}} - 8 \frac{1}{e^{2}} + 2 e^{- 2}$

Step 14.1.6

Combine $- 8$ and $\frac{1}{e^{2}}$ .

$\frac{4}{e^{2}} + \frac{- 8}{e^{2}} + 2 e^{- 2}$

Step 14.1.7

Move the negative in front of the fraction.

$\frac{4}{e^{2}} - \frac{8}{e^{2}} + 2 e^{- 2}$

Step 14.1.8

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

$\frac{4}{e^{2}} - \frac{8}{e^{2}} + 2 \frac{1}{e^{2}}$

Step 14.1.9

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

$\frac{4}{e^{2}} - \frac{8}{e^{2}} + \frac{2}{e^{2}}$

Step 14.2

Combine fractions.

Tap for more steps...

Step 14.2.1

Combine the numerators over the common denominator.

$\frac{4 - 8 + 2}{e^{2}}$

Step 14.2.2

Simplify the expression.

Tap for more steps...

Step 14.2.2.1

Subtract $8$ from $4$ .

$\frac{- 4 + 2}{e^{2}}$

Step 14.2.2.2

Add $- 4$ and $2$ .

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

Step 14.2.2.3

Move the negative in front of the fraction.

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

Step 15

$x = - 2$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 2$ is a local maximum

Step 16

Find the y-value when $x = - 2$ .

Tap for more steps...

Step 16.1

Replace the variable $x$ with $- 2$ in the expression.

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

Step 16.2

Simplify the result.

Tap for more steps...

Step 16.2.1

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

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

Step 16.2.2

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

$f (- 2) = 4 (\frac{1}{e^{2}})$

Step 16.2.3

Combine $4$ and $\frac{1}{e^{2}}$ .

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

Step 16.2.4

The final answer is $\frac{4}{e^{2}}$ .

$y = \frac{4}{e^{2}}$

Step 17

These are the local extrema for $f (x) = x^{2} e^{x}$ .

$(0, 0)$ is a local minima

$(- 2, \frac{4}{e^{2}})$ is a local maxima

Step 18