Calculus Examples

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

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

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

Find the first derivative of the function.

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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 $g (x) = x^{2}$ .

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To apply the Chain Rule, set $u$ as $x^{2}$ .

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

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

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

Replace all occurrences of $u$ with $x^{2}$ .

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

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

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

Simplify.

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Reorder the factors of $e^{x^{2}} (2 x)$ .

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

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

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

Find the second derivative of the function.

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

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

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

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

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 $g (x) = x^{2}$ .

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To apply the Chain Rule, set $u$ as $x^{2}$ .

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

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

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

Replace all occurrences of $u$ with $x^{2}$ .

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

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

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

Raise $x$ to the power of $1$ .

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

Raise $x$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Simplify the expression.

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Add $1$ and $1$ .

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

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

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

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

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

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

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

Simplify.

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Apply the distributive property.

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

Multiply $2$ by $2$ .

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

Reorder terms.

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

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

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

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

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

Find the first derivative.

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Find the first derivative.

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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 $g (x) = x^{2}$ .

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To apply the Chain Rule, set $u$ as $x^{2}$ .

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

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

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

Replace all occurrences of $u$ with $x^{2}$ .

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

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

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

Simplify.

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Reorder the factors of $e^{x^{2}} (2 x)$ .

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

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

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

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

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

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

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Set the first derivative equal to $0$ .

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

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

$x = 0$

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

Set $x$ equal to $0$ .

$x = 0$

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

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

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

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

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

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

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

Undefined

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

No solution

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

$x = 0$

Find the values where the derivative is undefined.

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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.

Critical points to evaluate.

$x = 0$

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.

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

Evaluate the second derivative.

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Simplify each term.

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

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

Multiply $4$ by $0$ .

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

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

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

Anything raised to $0$ is $1$ .

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

Multiply $0$ by $1$ .

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

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

$0 + 2 e^{0}$

Anything raised to $0$ is $1$ .

$0 + 2 \cdot 1$

Multiply $2$ by $1$ .

$0 + 2$

Add $0$ and $2$ .

$2$

$x = 0$ 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

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

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

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

Simplify the result.

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

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

Anything raised to $0$ is $1$ .

$f (0) = 1$

The final answer is $1$ .

$y = 1$

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

$(0, 1)$ is a local minima