Calculus Examples

$y = x + \frac{1}{x}$

Write $y = x + \frac{1}{x}$ as a function.

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

Find the first derivative of the function.

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Differentiate.

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By the Sum Rule, the derivative of $x + \frac{1}{x}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [\frac{1}{x}]$ .

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

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

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

Evaluate $\frac{d}{d x} [\frac{1}{x}]$ .

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

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

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

$1 - x^{- 2}$

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

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

Reorder terms.

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

Find the second derivative of the function.

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By the Sum Rule, the derivative of $- \frac{1}{x^{2}} + 1$ with respect to $x$ is $\frac{d}{d x} [- \frac{1}{x^{2}}] + \frac{d}{d x} [1]$ .

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

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

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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 $f (x) = - 1$ and $g (x) = \frac{1}{x^{2}}$ .

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

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

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

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 $x^{2}$ .

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

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

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

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

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

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

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

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$f'' (x) = - (- {(x^{2})}^{- 2} (2 x)) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} (1)$

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

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

$f'' (x) = - (- x^{2 \cdot - 2} (2 x)) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} (1)$

Multiply $2$ by $- 2$ .

$f'' (x) = - (- x^{- 4} (2 x)) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} (1)$

Multiply $2$ by $- 1$ .

$f'' (x) = - (- 2 x^{- 4} x) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} (1)$

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

$f'' (x) = - (- 2 (x \cdot x^{- 4})) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} (1)$

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

$f'' (x) = - (- 2 x^{1 - 4}) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} (1)$

Subtract $4$ from $1$ .

$f'' (x) = - (- 2 x^{- 3}) + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} (1)$

Multiply $- 2$ by $- 1$ .

$f'' (x) = 2 x^{- 3} + \frac{1}{x^{2}} \cdot 0 + \frac{d}{d x} (1)$

Multiply $\frac{1}{x^{2}}$ by $0$ .

$f'' (x) = 2 x^{- 3} + 0 + \frac{d}{d x} (1)$

Add $2 x^{- 3}$ and $0$ .

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

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

Simplify.

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

$f'' (x) = 2 (\frac{1}{x^{3}}) + 0$

Combine terms.

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Combine $2$ and $\frac{1}{x^{3}}$ .

$f'' (x) = \frac{2}{x^{3}} + 0$

Add $\frac{2}{x^{3}}$ and $0$ .

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

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

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

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 $x + \frac{1}{x}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [\frac{1}{x}]$ .

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

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

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

Evaluate $\frac{d}{d x} [\frac{1}{x}]$ .

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

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

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

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

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

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

Reorder terms.

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

The first derivative of $f (x)$ with respect to $x$ is $- \frac{1}{x^{2}} + 1$ .

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

Set the first derivative equal to $0$ then solve the equation $- \frac{1}{x^{2}} + 1 = 0$ .

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Set the first derivative equal to $0$ .

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

Subtract $1$ from both sides of the equation.

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

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.

$x^{2}, 1$

The LCM of one and any expression is the expression.

$x^{2}$

Multiply each term in $- \frac{1}{x^{2}} = - 1$ by $x^{2}$ to eliminate the fractions.

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Multiply each term in $- \frac{1}{x^{2}} = - 1$ by $x^{2}$ .

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

Simplify the left side.

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

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Move the leading negative in $- \frac{1}{x^{2}}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

$- 1 = - x^{2}$

Solve the equation.

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Rewrite the equation as $- x^{2} = - 1$ .

$- x^{2} = - 1$

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

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

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

Simplify the left side.

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Dividing two negative values results in a positive value.

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

Divide $x^{2}$ by $1$ .

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

Simplify the right side.

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Divide $- 1$ by $- 1$ .

$x^{2} = 1$

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

$x = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Find the values where the derivative is undefined.

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Set the denominator in $\frac{1}{x^{2}}$ 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.

$x = 1, - 1$

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.

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

Evaluate the second derivative.

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One to any power is one.

$\frac{2}{1}$

Divide $2$ by $1$ .

$2$

$x = 1$ 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.

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

Simplify the result.

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Divide $1$ by $1$ .

$f (1) = 1 + 1$

Add $1$ and $1$ .

$f (1) = 2$

The final answer is $2$ .

$y = 2$

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.

$\frac{2}{{(- 1)}^{3}}$

Evaluate the second derivative.

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Raise $- 1$ to the power of $3$ .

$\frac{2}{- 1}$

Divide $2$ by $- 1$ .

$- 2$

$x = - 1$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 1$ is a local maximum

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

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

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

Simplify the result.

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Divide $1$ by $- 1$ .

$f (- 1) = - 1 - 1$

Subtract $1$ from $- 1$ .

$f (- 1) = - 2$

The final answer is $- 2$ .

$y = - 2$

These are the local extrema for $f (x) = x + \frac{1}{x}$ .

$(1, 2)$ is a local minima

$(- 1, - 2)$ is a local maxima