Calculus Examples

$f (x, y) = x + \frac{4}{x} - y - \frac{9}{y} + 10$

Step 1

Write $f (x, y) - x - \frac{4}{x} + y + \frac{9}{y} - 10 = 0$ as a function.

$f (x) = f (x, y) - x - \frac{4}{x} + y + \frac{9}{y} - 10$

Step 2

Move all the expressions to the left side of the equation.

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Step 2.1

Subtract $x$ from both sides of the equation.

$f (x, y) - x = \frac{4}{x} - y - \frac{9}{y} + 10$

Step 2.2

Subtract $\frac{4}{x}$ from both sides of the equation.

$f (x, y) - x - \frac{4}{x} = - y - \frac{9}{y} + 10$

Step 2.3

Add $y$ to both sides of the equation.

$f (x, y) - x - \frac{4}{x} + y = - \frac{9}{y} + 10$

Step 2.4

Add $\frac{9}{y}$ to both sides of the equation.

$f (x, y) - x - \frac{4}{x} + y + \frac{9}{y} = 10$

Step 2.5

Subtract $10$ from both sides of the equation.

$f (x, y) - x - \frac{4}{x} + y + \frac{9}{y} - 10 = 0$

Step 3

Find the first derivative of the function.

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Step 3.1

By the Sum Rule, the derivative of $f (x, y) - x - \frac{4}{x} + y + \frac{9}{y} - 10$ with respect to $x$ is $\frac{d}{d x} [f (x, y)] + \frac{d}{d x} [- x] + \frac{d}{d x} [- \frac{4}{x}] + \frac{d}{d x} [y] + \frac{d}{d x} [\frac{9}{y}] + \frac{d}{d x} [- 10]$ .

$\frac{d}{d x} [f (x, y)] + \frac{d}{d x} [- x] + \frac{d}{d x} [- \frac{4}{x}] + \frac{d}{d x} [y] + \frac{d}{d x} [\frac{9}{y}] + \frac{d}{d x} [- 10]$

Step 3.2

Multiply $f$ by each element of the matrix.

$\frac{d}{d x} [(f x, f y)] + \frac{d}{d x} [- x] + \frac{d}{d x} [- \frac{4}{x}] + \frac{d}{d x} [y] + \frac{d}{d x} [\frac{9}{y}] + \frac{d}{d x} [- 10]$

Step 3.3

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

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Step 3.3.1

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

$\frac{d}{d x} [(f x, f y)] - \frac{d}{d x} [x] + \frac{d}{d x} [- \frac{4}{x}] + \frac{d}{d x} [y] + \frac{d}{d x} [\frac{9}{y}] + \frac{d}{d x} [- 10]$

Step 3.3.2

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

$\frac{d}{d x} [(f x, f y)] - 1 \cdot 1 + \frac{d}{d x} [- \frac{4}{x}] + \frac{d}{d x} [y] + \frac{d}{d x} [\frac{9}{y}] + \frac{d}{d x} [- 10]$

Step 3.3.3

Multiply $- 1$ by $1$ .

$\frac{d}{d x} [(f x, f y)] - 1 + \frac{d}{d x} [- \frac{4}{x}] + \frac{d}{d x} [y] + \frac{d}{d x} [\frac{9}{y}] + \frac{d}{d x} [- 10]$

Step 3.4

Evaluate $\frac{d}{d x} [- \frac{4}{x}]$ .

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Step 3.4.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- \frac{4}{x}$ with respect to $x$ is $- 4 \frac{d}{d x} [\frac{1}{x}]$ .

$\frac{d}{d x} [(f x, f y)] - 1 - 4 \frac{d}{d x} [\frac{1}{x}] + \frac{d}{d x} [y] + \frac{d}{d x} [\frac{9}{y}] + \frac{d}{d x} [- 10]$

Step 3.4.2

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

$\frac{d}{d x} [(f x, f y)] - 1 - 4 \frac{d}{d x} [x^{- 1}] + \frac{d}{d x} [y] + \frac{d}{d x} [\frac{9}{y}] + \frac{d}{d x} [- 10]$

Step 3.4.3

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

$\frac{d}{d x} [(f x, f y)] - 1 - 4 (- x^{- 2}) + \frac{d}{d x} [y] + \frac{d}{d x} [\frac{9}{y}] + \frac{d}{d x} [- 10]$

Step 3.4.4

Multiply $- 1$ by $- 4$ .

$\frac{d}{d x} [(f x, f y)] - 1 + 4 x^{- 2} + \frac{d}{d x} [y] + \frac{d}{d x} [\frac{9}{y}] + \frac{d}{d x} [- 10]$

Step 3.5

Differentiate using the Constant Rule.

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Step 3.5.1

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

$\frac{d}{d x} [(f x, f y)] - 1 + 4 x^{- 2} + 0 + \frac{d}{d x} [\frac{9}{y}] + \frac{d}{d x} [- 10]$

Step 3.5.2

Since $\frac{9}{y}$ is constant with respect to $x$ , the derivative of $\frac{9}{y}$ with respect to $x$ is $0$ .

$\frac{d}{d x} [(f x, f y)] - 1 + 4 x^{- 2} + 0 + 0 + \frac{d}{d x} [- 10]$

Step 3.5.3

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10$ with respect to $x$ is $0$ .

$\frac{d}{d x} [(f x, f y)] - 1 + 4 x^{- 2} + 0 + 0 + 0$

Step 3.6

Simplify.

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Step 3.6.1

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

$\frac{d}{d x} [(f x, f y)] - 1 + 4 \frac{1}{x^{2}} + 0 + 0 + 0$

Step 3.6.2

Combine terms.

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Step 3.6.2.1

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

$\frac{d}{d x} [(f x, f y)] - 1 + \frac{4}{x^{2}} + 0 + 0 + 0$

Step 3.6.2.2

Add $\frac{d}{d x} [(f x, f y)] - 1 + \frac{4}{x^{2}}$ and $0$ .

$\frac{d}{d x} [(f x, f y)] - 1 + \frac{4}{x^{2}} + 0 + 0$

Step 3.6.2.3

Add $\frac{d}{d x} [(f x, f y)] - 1 + \frac{4}{x^{2}}$ and $0$ .

$\frac{d}{d x} [(f x, f y)] - 1 + \frac{4}{x^{2}} + 0$

Step 3.6.2.4

Add $\frac{d}{d x} [(f x, f y)] - 1 + \frac{4}{x^{2}}$ and $0$ .

$\frac{d}{d x} [(f x, f y)] - 1 + \frac{4}{x^{2}}$

Step 3.6.3

Reorder terms.

$\frac{4}{x^{2}} + \frac{d}{d x} [(f x, f y)] - 1$

Step 4

Find the second derivative of the function.

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Step 4.1

By the Sum Rule, the derivative of $\frac{4}{x^{2}} + \frac{d}{d x} ((f x, f y)) - 1$ with respect to $x$ is $\frac{d}{d x} [\frac{4}{x^{2}}] + \frac{d}{d x} [\frac{d}{d x} ((f x, f y))] + \frac{d}{d x} [- 1]$ .

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

Step 4.2

Evaluate $\frac{d}{d x} [\frac{4}{x^{2}}]$ .

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Step 4.2.1

Since $4$ is constant with respect to $x$ , the derivative of $\frac{4}{x^{2}}$ with respect to $x$ is $4 \frac{d}{d x} [\frac{1}{x^{2}}]$ .

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

Step 4.2.2

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

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

Step 4.2.3

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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Step 4.2.3.1

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

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

Step 4.2.3.2

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

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

Step 4.2.3.3

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

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

Step 4.2.4

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

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

Step 4.2.5

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

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Step 4.2.5.1

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

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

Step 4.2.5.2

Multiply $2$ by $- 2$ .

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

Step 4.2.6

Multiply $2$ by $- 1$ .

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

Step 4.2.7

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

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

Step 4.2.8

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

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

Step 4.2.9

Subtract $4$ from $1$ .

$f'' (x) = 4 (- 2 x^{- 3}) + \frac{d}{d x} (\frac{d}{d x} \cdot ((f x, f y))) + \frac{d}{d x} (- 1)$

Step 4.2.10

Multiply $- 2$ by $4$ .

$f'' (x) = - 8 x^{- 3} + \frac{d}{d x} (\frac{d}{d x} \cdot ((f x, f y))) + \frac{d}{d x} (- 1)$

Step 4.3

Evaluate $\frac{d}{d x} [\frac{d}{d x} ((f x, f y))]$ .

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Step 4.3.1

Cancel the common factor of $d$ .

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Step 4.3.1.1

Cancel the common factor.

$f'' (x) = - 8 x^{- 3} + \frac{d}{d x} (\frac{d}{d x} \cdot (f x, f y)) + \frac{d}{d x} (- 1)$

Step 4.3.1.2

Rewrite the expression.

$f'' (x) = - 8 x^{- 3} + \frac{d}{d x} (\frac{1}{x} \cdot (f x, f y)) + \frac{d}{d x} (- 1)$

Step 4.3.2

Multiply $\frac{1}{x}$ by each element of the matrix.

$f'' (x) = - 8 x^{- 3} + \frac{d}{d x} ((\frac{1}{x} \cdot (f x), \frac{1}{x} \cdot (f y))) + \frac{d}{d x} (- 1)$

Step 4.3.3

Simplify each element in the matrix.

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Step 4.3.3.1

Cancel the common factor of $x$ .

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Step 4.3.3.1.1

Factor $x$ out of $f x$ .

$f'' (x) = - 8 x^{- 3} + \frac{d}{d x} ((\frac{1}{x} \cdot (x f), \frac{1}{x} \cdot (f y))) + \frac{d}{d x} (- 1)$

Step 4.3.3.1.2

Cancel the common factor.

$f'' (x) = - 8 x^{- 3} + \frac{d}{d x} ((\frac{1}{x} \cdot (x f), \frac{1}{x} \cdot (f y))) + \frac{d}{d x} (- 1)$

Step 4.3.3.1.3

Rewrite the expression.

$f'' (x) = - 8 x^{- 3} + \frac{d}{d x} ((f, \frac{1}{x} \cdot (f y))) + \frac{d}{d x} (- 1)$

Step 4.3.3.2

Multiply $\frac{1}{x} (f y)$ .

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Step 4.3.3.2.1

Combine $f$ and $\frac{1}{x}$ .

$f'' (x) = - 8 x^{- 3} + \frac{d}{d x} ((f, \frac{f}{x} \cdot y)) + \frac{d}{d x} (- 1)$

Step 4.3.3.2.2

Combine $\frac{f}{x}$ and $y$ .

$f'' (x) = - 8 x^{- 3} + \frac{d}{d x} ((f, \frac{f y}{x})) + \frac{d}{d x} (- 1)$

Step 4.4

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

$f'' (x) = - 8 x^{- 3} + \frac{d}{d x} ((f, \frac{f y}{x})) + 0$

Step 4.5

Simplify.

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Step 4.5.1

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

$f'' (x) = - 8 \frac{1}{x^{3}} + \frac{d}{d x} ((f, \frac{f y}{x})) + 0$

Step 4.5.2

Combine terms.

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Step 4.5.2.1

Combine $- 8$ and $\frac{1}{x^{3}}$ .

$f'' (x) = \frac{- 8}{x^{3}} + \frac{d}{d x} ((f, \frac{f y}{x})) + 0$

Step 4.5.2.2

Move the negative in front of the fraction.

$f'' (x) = - \frac{8}{x^{3}} + \frac{d}{d x} ((f, \frac{f y}{x})) + 0$

Step 4.5.2.3

Add $- \frac{8}{x^{3}} + \frac{d}{d x} [(f, \frac{f y}{x})]$ and $0$ .

$f'' (x) = - \frac{8}{x^{3}} + \frac{d}{d x} ((f, \frac{f y}{x}))$

Step 5

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

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

Step 6

Since there is no value of $x$ that makes the first derivative equal to $0$ , there are no local extrema.

No Local Extrema

Step 7

No Local Extrema

Step 8