Calculus Examples

$f (x) = x e^{x + y}$

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

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

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 $g (x) = x + y$ .

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

Replace all occurrences of $u$ with $x + y$ .

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

Step 1.3

Differentiate.

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

Since $y$ is constant with respect to $x$ , the derivative of $y$ with respect to $x$ is $0$ .

$x (e^{x + y} (1 + 0)) + e^{x + y} \frac{d}{d x} [x]$

Step 1.3.4

Simplify the expression.

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Step 1.3.4.1

Add $1$ and $0$ .

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

Step 1.3.4.2

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

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

Step 1.3.5

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

$x e^{x + y} + e^{x + y} \cdot 1$

Step 1.3.6

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

$x e^{x + y} + e^{x + y}$

Step 2

Find the second derivative of the function.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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 $g (x) = x + y$ .

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Step 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Since $y$ is constant with respect to $x$ , the derivative of $y$ with respect to $x$ is $0$ .

$f'' (x) = x (e^{x + y} (1 + 0)) + e^{x + y} \frac{d}{d x} (x) + \frac{d}{d x} (e^{x + y})$

Step 2.2.6

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

$f'' (x) = x (e^{x + y} (1 + 0)) + e^{x + y} \cdot 1 + \frac{d}{d x} (e^{x + y})$

Step 2.2.7

Add $1$ and $0$ .

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

Step 2.2.8

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

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

Step 2.2.9

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

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

Step 2.3

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

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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 $g (x) = x + y$ .

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Step 2.3.1.1

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

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

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 $a$ = $e$ .

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

Step 2.3.1.3

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Since $y$ is constant with respect to $x$ , the derivative of $y$ with respect to $x$ is $0$ .

$f'' (x) = x e^{x + y} + e^{x + y} + e^{x + y} (1 + 0)$

Step 2.3.5

Add $1$ and $0$ .

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

Step 2.3.6

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

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

Step 2.4

Simplify.

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Step 2.4.1

Add $e^{x + y}$ and $e^{x + y}$ .

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

Step 2.4.2

Reorder terms.

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

Step 2.4.3

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

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

Step 3

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

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

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

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

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 $g (x) = x + y$ .

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Step 4.1.2.1

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

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

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 $a$ = $e$ .

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

Step 4.1.2.3

Replace all occurrences of $u$ with $x + y$ .

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

Step 4.1.3

Differentiate.

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Step 4.1.3.1

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

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

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 $n = 1$ .

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

Step 4.1.3.3

Since $y$ is constant with respect to $x$ , the derivative of $y$ with respect to $x$ is $0$ .

$x (e^{x + y} (1 + 0)) + e^{x + y} \frac{d}{d x} [x]$

Step 4.1.3.4

Simplify the expression.

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Step 4.1.3.4.1

Add $1$ and $0$ .

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

Step 4.1.3.4.2

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

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

Step 4.1.3.5

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

$x e^{x + y} + e^{x + y} \cdot 1$

Step 4.1.3.6

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

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

Step 4.2

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

$x e^{x + y} + e^{x + y}$

Step 5

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

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Step 5.1

Set the first derivative equal to $0$ .

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

Multiply by $1$ .

$e^{x + y} x + e^{x + y} \cdot 1 = 0$

Step 5.2.3

Factor $e^{x + y}$ out of $e^{x + y} x + e^{x + y} \cdot 1$ .

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

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 $0$ .

$e^{x + y} = 0$

$x + 1 = 0$

Step 5.4

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

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Step 5.4.1

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

$e^{x + y} = 0$

Step 5.4.2

Solve $e^{x + y} = 0$ 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.

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

Step 5.4.2.2

Expand the left side.

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Step 5.4.2.2.1

Expand $\ln (e^{x + y})$ by moving $x + y$ outside the logarithm.

$(x + y) \ln (e) = \ln (0)$

Step 5.4.2.2.2

The natural logarithm of $e$ is $1$ .

$(x + y) \cdot 1 = \ln (0)$

Step 5.4.2.2.3

Multiply $x + y$ by $1$ .

$x + y = \ln (0)$

Step 5.4.2.3

Simplify the right side.

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Step 5.4.2.3.1

The equation cannot be solved because it is undefined.

$Undefined$

Step 5.4.2.4

Subtract $y$ from both sides of the equation.

$x = Undefined - y$

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

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5.6

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

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

Step 9

Evaluate the second derivative.

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Step 9.1

Rewrite $- 1 e^{- 1 + y}$ as $- e^{- 1 + y}$ .

$- e^{- 1 + y} + 2 e^{- 1 + y}$

Step 9.2

Add $- e^{- 1 + y}$ and $2 e^{- 1 + y}$ .

$e^{- 1 + y}$

Step 10

Since the first derivative test failed, there are no local extrema.

No Local Extrema

Step 11