Calculus Examples

$x \ln (y) + y \ln (x) = 2$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x \ln (y) + y \ln (x)) = \frac{d}{d x} (2)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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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 $g (x) = \ln (y)$ .

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

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 $f (x) = \ln (x)$ and $g (x) = y$ .

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Step 2.2.2.1

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

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

Step 2.2.2.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$x (\frac{1}{u} \frac{d}{d x} [y]) + \ln (y) \frac{d}{d x} [x] + \frac{d}{d x} [y \ln (x)]$

Step 2.2.2.3

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

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

Step 2.2.3

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$x (\frac{1}{y} y') + \ln (y) \frac{d}{d x} [x] + \frac{d}{d x} [y \ln (x)]$

Step 2.2.4

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

$x (\frac{1}{y} y') + \ln (y) \cdot 1 + \frac{d}{d x} [y \ln (x)]$

Step 2.2.5

Combine $\frac{1}{y}$ and $y'$ .

$x \frac{y'}{y} + \ln (y) \cdot 1 + \frac{d}{d x} [y \ln (x)]$

Step 2.2.6

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

$\frac{x y'}{y} + \ln (y) \cdot 1 + \frac{d}{d x} [y \ln (x)]$

Step 2.2.7

Multiply $\ln (y)$ by $1$ .

$\frac{x y'}{y} + \ln (y) + \frac{d}{d x} [y \ln (x)]$

Step 2.3

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

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Step 2.3.1

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 $f (x) = y$ and $g (x) = \ln (x)$ .

$\frac{x y'}{y} + \ln (y) + y \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [y]$

Step 2.3.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

$\frac{x y'}{y} + \ln (y) + y \frac{1}{x} + \ln (x) \frac{d}{d x} [y]$

Step 2.3.3

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$\frac{x y'}{y} + \ln (y) + y \frac{1}{x} + \ln (x) y'$

Step 2.3.4

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

$\frac{x y'}{y} + \ln (y) + \frac{y}{x} + \ln (x) y'$

Step 2.4

Reorder terms.

$\ln (x) y' + \frac{x y'}{y} + \ln (y) + \frac{y}{x}$

Step 3

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$0$

Step 4

Reform the equation by setting the left side equal to the right side.

$\ln (x) y' + \frac{x y'}{y} + \ln (y) + \frac{y}{x} = 0$

Step 5

Solve for $y'$ .

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Step 5.1

Multiply each term in $\ln (x) y' + \frac{x y'}{y} + \ln (y) + \frac{y}{x} = 0$ by $y x$ to eliminate the fractions.

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Step 5.1.1

Multiply each term in $\ln (x) y' + \frac{x y'}{y} + \ln (y) + \frac{y}{x} = 0$ by $y x$ .

$\ln (x) y' (y x) + \frac{x y'}{y} (y x) + \ln (y) (y x) + \frac{y}{x} (y x) = 0 (y x)$

Step 5.1.2

Simplify the left side.

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Step 5.1.2.1

Simplify each term.

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Step 5.1.2.1.1

Cancel the common factor of $y$ .

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Step 5.1.2.1.1.1

Factor $y$ out of $y x$ .

$\ln (x) y' y x + \frac{x y'}{y} (y (x)) + \ln (y) (y x) + \frac{y}{x} (y x) = 0 (y x)$

Step 5.1.2.1.1.2

Cancel the common factor.

$\ln (x) y' y x + \frac{x y'}{y} (y x) + \ln (y) (y x) + \frac{y}{x} (y x) = 0 (y x)$

Step 5.1.2.1.1.3

Rewrite the expression.

$\ln (x) y' y x + x y' x + \ln (y) (y x) + \frac{y}{x} (y x) = 0 (y x)$

Step 5.1.2.1.2

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

$\ln (x) y' y x + y' (x^{1} x) + \ln (y) (y x) + \frac{y}{x} (y x) = 0 (y x)$

Step 5.1.2.1.3

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

$\ln (x) y' y x + y' (x^{1} x^{1}) + \ln (y) (y x) + \frac{y}{x} (y x) = 0 (y x)$

Step 5.1.2.1.4

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

$\ln (x) y' y x + y' x^{1 + 1} + \ln (y) (y x) + \frac{y}{x} (y x) = 0 (y x)$

Step 5.1.2.1.5

Add $1$ and $1$ .

$\ln (x) y' y x + y' x^{2} + \ln (y) (y x) + \frac{y}{x} (y x) = 0 (y x)$

Step 5.1.2.1.6

Cancel the common factor of $x$ .

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Step 5.1.2.1.6.1

Factor $x$ out of $y x$ .

$\ln (x) y' y x + y' x^{2} + \ln (y) y x + \frac{y}{x} (x y) = 0 (y x)$

Step 5.1.2.1.6.2

Cancel the common factor.

$\ln (x) y' y x + y' x^{2} + \ln (y) y x + \frac{y}{x} (x y) = 0 (y x)$

Step 5.1.2.1.6.3

Rewrite the expression.

$\ln (x) y' y x + y' x^{2} + \ln (y) y x + y \cdot y = 0 (y x)$

Step 5.1.2.1.7

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

$\ln (x) y' y x + y' x^{2} + \ln (y) y x + y^{1} y = 0 (y x)$

Step 5.1.2.1.8

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

$\ln (x) y' y x + y' x^{2} + \ln (y) y x + y^{1} y^{1} = 0 (y x)$

Step 5.1.2.1.9

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

$\ln (x) y' y x + y' x^{2} + \ln (y) y x + y^{1 + 1} = 0 (y x)$

Step 5.1.2.1.10

Add $1$ and $1$ .

$\ln (x) y' y x + y' x^{2} + \ln (y) y x + y^{2} = 0 (y x)$

Step 5.1.3

Simplify the right side.

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Step 5.1.3.1

Multiply $0 (y x)$ .

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Step 5.1.3.1.1

Multiply $y$ by $0$ .

$\ln (x) y' y x + y' x^{2} + \ln (y) y x + y^{2} = 0 x$

Step 5.1.3.1.2

Multiply $0$ by $x$ .

$\ln (x) y' y x + y' x^{2} + \ln (y) y x + y^{2} = 0$

Step 5.2

Move all the terms containing a logarithm to the left side of the equation.

$\ln (x) y' y x + \ln (y) y x = - y' x^{2} - y^{2} + 0$

Step 5.3

Reorder factors in $\ln (x) y' y x + \ln (y) y x$ .

$y' y x \ln (x) + y x \ln (y) = - y' x^{2} - y^{2} + 0$

Step 5.4

Add $- y' x^{2} - y^{2}$ and $0$ .

$y' y x \ln (x) + y x \ln (y) = - y' x^{2} - y^{2}$

Step 5.5

Add $y' x^{2}$ to both sides of the equation.

$y' y x \ln (x) + y x \ln (y) + y' x^{2} = - y^{2}$

Step 5.6

Subtract $y x \ln (y)$ from both sides of the equation.

$y' y x \ln (x) + y' x^{2} = - y^{2} - y x \ln (y)$

Step 5.7

Factor $y' x$ out of $y' y x \ln (x) + y' x^{2}$ .

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Step 5.7.1

Factor $y' x$ out of $y' y x \ln (x)$ .

$y' x (y \ln (x)) + y' x^{2} = - y^{2} - y x \ln (y)$

Step 5.7.2

Factor $y' x$ out of $y' x^{2}$ .

$y' x (y \ln (x)) + y' x \cdot x = - y^{2} - y x \ln (y)$

Step 5.7.3

Factor $y' x$ out of $y' x (y \ln (x)) + y' x \cdot x$ .

$y' x (y \ln (x) + x) = - y^{2} - y x \ln (y)$

Step 5.8

Divide each term in $y' x (y \ln (x) + x) = - y^{2} - y x \ln (y)$ by $x (y \ln (x) + x)$ and simplify.

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Step 5.8.1

Divide each term in $y' x (y \ln (x) + x) = - y^{2} - y x \ln (y)$ by $x (y \ln (x) + x)$ .

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

Step 5.8.2

Simplify the left side.

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Step 5.8.2.1

Cancel the common factor of $x$ .

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Step 5.8.2.1.1

Cancel the common factor.

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

Step 5.8.2.1.2

Rewrite the expression.

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

Step 5.8.2.2

Cancel the common factor of $y \ln (x) + x$ .

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Step 5.8.2.2.1

Cancel the common factor.

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

Step 5.8.2.2.2

Divide $y'$ by $1$ .

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

Step 5.8.3

Simplify the right side.

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Step 5.8.3.1

Simplify each term.

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Step 5.8.3.1.1

Move the negative in front of the fraction.

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

Step 5.8.3.1.2

Cancel the common factor of $x$ .

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Step 5.8.3.1.2.1

Cancel the common factor.

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

Step 5.8.3.1.2.2

Rewrite the expression.

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

Step 5.8.3.1.3

Move the negative in front of the fraction.

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

Step 5.8.3.2

To write $- \frac{y \ln (y)}{y \ln (x) + x}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$y' = - \frac{y^{2}}{x (y \ln (x) + x)} - \frac{y \ln (y)}{y \ln (x) + x} \cdot \frac{x}{x}$

Step 5.8.3.3

Write each expression with a common denominator of $x (y \ln (x) + x)$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.8.3.3.1

Multiply $\frac{y \ln (y)}{y \ln (x) + x}$ by $\frac{x}{x}$ .

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

Step 5.8.3.3.2

Reorder the factors of $(y \ln (x) + x) x$ .

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

Step 5.8.3.4

Combine the numerators over the common denominator.

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

Step 5.8.3.5

Simplify the numerator.

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Step 5.8.3.5.1

Factor $y$ out of $- y^{2} - y \ln (y) x$ .

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Step 5.8.3.5.1.1

Factor $y$ out of $- y^{2}$ .

$y' = \frac{y (- y) - y \ln (y) x}{x (y \ln (x) + x)}$

Step 5.8.3.5.1.2

Factor $y$ out of $- y \ln (y) x$ .

$y' = \frac{y (- y) + y (- 1 \ln (y) x)}{x (y \ln (x) + x)}$

Step 5.8.3.5.1.3

Factor $y$ out of $y (- y) + y (- 1 \ln (y) x)$ .

$y' = \frac{y (- y - 1 \ln (y) x)}{x (y \ln (x) + x)}$

Step 5.8.3.5.2

Rewrite $- 1 \ln (y)$ as $- \ln (y)$ .

$y' = \frac{y (- y - \ln (y) x)}{x (y \ln (x) + x)}$

Step 5.8.3.6

Simplify with factoring out.

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Step 5.8.3.6.1

Factor $- 1$ out of $- y$ .

$y' = \frac{y (- (y) - \ln (y) x)}{x (y \ln (x) + x)}$

Step 5.8.3.6.2

Factor $- 1$ out of $- \ln (y) x$ .

$y' = \frac{y (- (y) - (\ln (y) x))}{x (y \ln (x) + x)}$

Step 5.8.3.6.3

Factor $- 1$ out of $- (y) - (\ln (y) x)$ .

$y' = \frac{y (- (y + \ln (y) x))}{x (y \ln (x) + x)}$

Step 5.8.3.6.4

Simplify the expression.

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Step 5.8.3.6.4.1

Rewrite $- (y + \ln (y) x)$ as $- 1 (y + \ln (y) x)$ .

$y' = \frac{y (- 1 (y + \ln (y) x))}{x (y \ln (x) + x)}$

Step 5.8.3.6.4.2

Move the negative in front of the fraction.

$y' = - \frac{y (y + \ln (y) x)}{x (y \ln (x) + x)}$

Step 6

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = - \frac{y (y + \ln (y) x)}{x (y \ln (x) + x)}$