Calculus Examples

$x^{2} + \frac{2}{x}$

Find the first derivative of the function.

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Differentiate.

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

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

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

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

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

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Since $2$ is constant with respect to $x$ , the derivative of $\frac{2}{x}$ with respect to $x$ is $2 \frac{d}{d x} [\frac{1}{x}]$ .

$2 x + 2 \frac{d}{d x} [\frac{1}{x}]$

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$2 x + 2 \frac{d}{d x} [x^{- 1}]$

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

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

Multiply $- 1$ by $2$ .

$2 x - 2 x^{- 2}$

Simplify.

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

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

Combine terms.

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

$2 x + \frac{- 2}{x^{2}}$

Move the negative in front of the fraction.

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

Find the second derivative of the function.

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

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

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Since $- 2$ is constant with respect to $x$ , the derivative of $- \frac{2}{x^{2}}$ with respect to $x$ is $- 2 \frac{d}{d x} [\frac{1}{x^{2}}]$ .

$f'' (x) = 2 - 2 \frac{d}{d x} \frac{1}{x^{2}}$

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

$f'' (x) = 2 - 2 \frac{d}{d x} {(x^{2})}^{- 1}$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2}$ .

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

$f'' (x) = 2 - 2 (\frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x^{2}))$

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

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

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

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

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

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

Multiply the exponents in ${(x^{2})}^{- 2}$ .

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Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f'' (x) = 2 - 2 (- x^{2 \cdot - 2} (2 x))$

Multiply $2$ by $- 2$ .

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

Multiply $2$ by $- 1$ .

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

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

$f'' (x) = 2 - 2 (- 2 (x \cdot x^{- 4}))$

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

$f'' (x) = 2 - 2 (- 2 x^{1 - 4})$

Subtract $4$ from $1$ .

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

Multiply $- 2$ by $- 2$ .

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

Simplify.

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

$f'' (x) = 2 + 4 (\frac{1}{x^{3}})$

Combine $4$ and $\frac{1}{x^{3}}$ .

$f'' (x) = 2 + \frac{4}{x^{3}}$

Reorder terms.

$f'' (x) = \frac{4}{x^{3}} + 2$

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

$2 x - \frac{2}{x^{2}} = 0$

Find the first derivative.

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Find the first derivative.

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Differentiate.

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

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

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

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

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

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Since $2$ is constant with respect to $x$ , the derivative of $\frac{2}{x}$ with respect to $x$ is $2 \frac{d}{d x} [\frac{1}{x}]$ .

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

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

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

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

Multiply $- 1$ by $2$ .

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

Simplify.

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

$f' (x) = 2 x - 2 \frac{1}{x^{2}}$

Combine terms.

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

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

Move the negative in front of the fraction.

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

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

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

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

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Set the first derivative equal to $0$ .

$2 x - \frac{2}{x^{2}} = 0$

Find the LCD of the terms in the equation.

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Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, x^{2}, 1$

The LCM of one and any expression is the expression.

$x^{2}$

Multiply each term in $2 x - \frac{2}{x^{2}} = 0$ by $x^{2}$ to eliminate the fractions.

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Multiply each term in $2 x - \frac{2}{x^{2}} = 0$ by $x^{2}$ .

$2 x \cdot x^{2} - \frac{2}{x^{2}} x^{2} = 0 x^{2}$

Simplify the left side.

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Simplify each term.

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Multiply $x$ by $x^{2}$ by adding the exponents.

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Move $x^{2}$ .

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

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

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

$2 (x^{2} x^{1}) - \frac{2}{x^{2}} x^{2} = 0 x^{2}$

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

$2 x^{2 + 1} - \frac{2}{x^{2}} x^{2} = 0 x^{2}$

Add $2$ and $1$ .

$2 x^{3} - \frac{2}{x^{2}} x^{2} = 0 x^{2}$

Cancel the common factor of $x^{2}$ .

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Move the leading negative in $- \frac{2}{x^{2}}$ into the numerator.

$2 x^{3} + \frac{- 2}{x^{2}} x^{2} = 0 x^{2}$

Cancel the common factor.

$2 x^{3} + \frac{- 2}{x^{2}} x^{2} = 0 x^{2}$

Rewrite the expression.

$2 x^{3} - 2 = 0 x^{2}$

Simplify the right side.

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Multiply $0$ by $x^{2}$ .

$2 x^{3} - 2 = 0$

Solve the equation.

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Add $2$ to both sides of the equation.

$2 x^{3} = 2$

Subtract $2$ from both sides of the equation.

$2 x^{3} - 2 = 0$

Factor the left side of the equation.

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

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Factor $2$ out of $2 x^{3}$ .

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

Factor $2$ out of $- 2$ .

$2 (x^{3}) + 2 (- 1) = 0$

Factor $2$ out of $2 (x^{3}) + 2 (- 1)$ .

$2 (x^{3} - 1) = 0$

Rewrite $1$ as $1^{3}$ .

$2 (x^{3} - 1^{3}) = 0$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 1$ .

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

Factor.

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Simplify.

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Multiply $x$ by $1$ .

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

One to any power is one.

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

Remove unnecessary parentheses.

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

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

$x - 1 = 0$

$x^{2} + x + 1 = 0$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x^{2} + x + 1$ equal to $0$ and solve for $x$ .

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Set $x^{2} + x + 1$ equal to $0$ .

$x^{2} + x + 1 = 0$

Solve $x^{2} + x + 1 = 0$ for $x$ .

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Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Simplify.

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Simplify the numerator.

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One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

$x = \frac{- 1 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{- 1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

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

$x = 1, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Find the values where the derivative is undefined.

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Set the denominator in $\frac{2}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Solve for $x$ .

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

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Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Critical points to evaluate.

$x = 1$

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.

$\frac{4}{{(1)}^{3}} + 2$

Evaluate the second derivative.

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Simplify each term.

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One to any power is one.

$\frac{4}{1} + 2$

Divide $4$ by $1$ .

$4 + 2$

Add $4$ and $2$ .

$6$

$x = 1$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 1$ is a local minimum

Find the y-value when $x = 1$ .

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Replace the variable $x$ with $1$ in the expression.

$f (1) = {(1)}^{2} + \frac{2}{1}$

Simplify the result.

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Simplify each term.

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One to any power is one.

$f (1) = 1 + \frac{2}{1}$

Divide $2$ by $1$ .

$f (1) = 1 + 2$

Add $1$ and $2$ .

$f (1) = 3$

The final answer is $3$ .

$y = 3$

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

$(1, 3)$ is a local minima