Calculus Examples

$1 + \ln (x y) = e^{x - y}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (1 + \ln (x y)) = \frac{d}{d x} (e^{x - y})$

Step 2

Differentiate the left side of the equation.

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Step 2.1

Differentiate.

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Step 2.1.1

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

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

Step 2.1.2

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

$0 + \frac{d}{d x} [\ln (x y)]$

Step 2.2

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

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Step 2.2.1

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

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Step 2.2.1.1

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

$0 + \frac{d}{d u} [\ln (u)] \frac{d}{d x} [x y]$

Step 2.2.1.2

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

$0 + \frac{1}{u} \frac{d}{d x} [x y]$

Step 2.2.1.3

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

$0 + \frac{1}{x y} \frac{d}{d x} [x y]$

Step 2.2.2

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 $g (x) = y$ .

$0 + \frac{1}{x y} (x \frac{d}{d x} [y] + y \frac{d}{d x} [x])$

Step 2.2.3

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

$0 + \frac{1}{x y} (x y' + y \frac{d}{d x} [x])$

Step 2.2.4

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

$0 + \frac{1}{x y} (x y' + y \cdot 1)$

Step 2.2.5

Multiply $y$ by $1$ .

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

Step 2.3

Simplify.

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Step 2.3.1

Apply the distributive property.

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

Step 2.3.2

Combine terms.

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Step 2.3.2.1

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

$0 + \frac{x}{x y} y' + \frac{1}{x y} y$

Step 2.3.2.2

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

$0 + \frac{x y'}{x y} + \frac{1}{x y} y$

Step 2.3.2.3

Cancel the common factor of $x$ .

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Step 2.3.2.3.1

Cancel the common factor.

$0 + \frac{x y'}{x y} + \frac{1}{x y} y$

Step 2.3.2.3.2

Rewrite the expression.

$0 + \frac{y'}{y} + \frac{1}{x y} y$

Step 2.3.2.4

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

$0 + \frac{y'}{y} + \frac{y}{x y}$

Step 2.3.2.5

Cancel the common factor of $y$ .

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Step 2.3.2.5.1

Cancel the common factor.

$0 + \frac{y'}{y} + \frac{y}{x y}$

Step 2.3.2.5.2

Rewrite the expression.

$0 + \frac{y'}{y} + \frac{1}{x}$

Step 2.3.2.6

Add $0$ and $\frac{y'}{y} + \frac{1}{x}$ .

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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Step 3.1.1

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

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

Step 3.1.2

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

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

Step 3.1.3

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

$e^{x - y} \frac{d}{d x} [x - y]$

Step 3.2

Differentiate.

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Step 3.2.1

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

$e^{x - y} (\frac{d}{d x} [x] + \frac{d}{d x} [- y])$

Step 3.2.2

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

$e^{x - y} (1 + \frac{d}{d x} [- y])$

Step 3.2.3

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

$e^{x - y} (1 - \frac{d}{d x} [y])$

Step 3.3

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

$e^{x - y} (1 - y')$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Simplify $e^{x - y} (1 - y')$ .

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Step 5.1.1

Rewrite.

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

Step 5.1.2

Simplify by adding zeros.

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

Step 5.1.3

Apply the distributive property.

$\frac{y'}{y} + \frac{1}{x} = e^{x - y} \cdot 1 + e^{x - y} (- y')$

Step 5.1.4

Simplify the expression.

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Step 5.1.4.1

Multiply $e^{x - y}$ by $1$ .

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

Step 5.1.4.2

Rewrite using the commutative property of multiplication.

$\frac{y'}{y} + \frac{1}{x} = e^{x - y} - e^{x - y} y'$

Step 5.1.4.3

Reorder factors in $e^{x - y} - e^{x - y} y'$ .

$\frac{y'}{y} + \frac{1}{x} = e^{x - y} - y' e^{x - y}$

Step 5.2

Move all terms containing $y'$ to the left side of the equation.

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Step 5.2.1

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

$\frac{y'}{y} + \frac{1}{x} + y' e^{x - y} = e^{x - y}$

Step 5.2.2

To write $y' e^{x - y}$ as a fraction with a common denominator, multiply by $\frac{y}{y}$ .

$\frac{y'}{y} + \frac{y' e^{x - y} y}{y} + \frac{1}{x} = e^{x - y}$

Step 5.2.3

Combine the numerators over the common denominator.

$\frac{y' + y' e^{x - y} y}{y} + \frac{1}{x} = e^{x - y}$

Step 5.2.4

Factor $y'$ out of $y' + y' e^{x - y} y$ .

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Step 5.2.4.1

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

$\frac{{y'}^{1} + y' e^{x - y} y}{y} + \frac{1}{x} = e^{x - y}$

Step 5.2.4.2

Factor $y'$ out of ${y'}^{1}$ .

$\frac{y' \cdot 1 + y' e^{x - y} y}{y} + \frac{1}{x} = e^{x - y}$

Step 5.2.4.3

Factor $y'$ out of $y' e^{x - y} y$ .

$\frac{y' \cdot 1 + y' (e^{x - y} y)}{y} + \frac{1}{x} = e^{x - y}$

Step 5.2.4.4

Factor $y'$ out of $y' \cdot 1 + y' (e^{x - y} y)$ .

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

Step 5.2.5

To write $\frac{y' (1 + e^{x - y} y)}{y}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$\frac{y' (1 + e^{x - y} y)}{y} \cdot \frac{x}{x} + \frac{1}{x} = e^{x - y}$

Step 5.2.6

To write $\frac{1}{x}$ as a fraction with a common denominator, multiply by $\frac{y}{y}$ .

$\frac{y' (1 + e^{x - y} y)}{y} \cdot \frac{x}{x} + \frac{1}{x} \cdot \frac{y}{y} = e^{x - y}$

Step 5.2.7

Write each expression with a common denominator of $y x$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.2.7.1

Multiply $\frac{y' (1 + e^{x - y} y)}{y}$ by $\frac{x}{x}$ .

$\frac{y' (1 + e^{x - y} y) x}{y x} + \frac{1}{x} \cdot \frac{y}{y} = e^{x - y}$

Step 5.2.7.2

Multiply $\frac{1}{x}$ by $\frac{y}{y}$ .

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

Step 5.2.7.3

Reorder the factors of $y x$ .

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

Step 5.2.8

Combine the numerators over the common denominator.

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

Step 5.2.9

Simplify the numerator.

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Step 5.2.9.1

Apply the distributive property.

$\frac{(y' \cdot 1 + y' (e^{x - y} y)) x + y}{x y} = e^{x - y}$

Step 5.2.9.2

Multiply $y'$ by $1$ .

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

Step 5.2.9.3

Apply the distributive property.

$\frac{y' x + y' e^{x - y} y x + y}{x y} = e^{x - y}$

Step 5.2.10

Reorder factors in $\frac{y' x + y' e^{x - y} y x + y}{x y}$ .

$\frac{y' x + y' y x e^{x - y} + y}{x y} = e^{x - y}$

Step 5.3

Multiply both sides by $x y$ .

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

Step 5.4

Simplify.

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Step 5.4.1

Simplify the left side.

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Step 5.4.1.1

Simplify $\frac{y' x + y' y x e^{x - y} + y}{x y} (x y)$ .

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Step 5.4.1.1.1

Cancel the common factor of $x y$ .

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Step 5.4.1.1.1.1

Cancel the common factor.

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

Step 5.4.1.1.1.2

Rewrite the expression.

$y' x + y' y x e^{x - y} + y = e^{x - y} (x y)$

Step 5.4.1.1.2

Reorder.

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Step 5.4.1.1.2.1

Move $y$ .

$y' x + y' x y e^{x - y} + y = e^{x - y} (x y)$

Step 5.4.1.1.2.2

Move $y' x y e^{x - y}$ .

$y' x + y + y' x y e^{x - y} = e^{x - y} (x y)$

Step 5.4.2

Simplify the right side.

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Step 5.4.2.1

Reorder factors in $e^{x - y} (x y)$ .

$y' x + y + y' x y e^{x - y} = x y e^{x - y}$

Step 5.5

Solve for $y'$ .

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Step 5.5.1

Subtract $y$ from both sides of the equation.

$y' x + y' x y e^{x - y} = x y e^{x - y} - y$

Step 5.5.2

Factor $y' x$ out of $y' x + y' x y e^{x - y}$ .

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Step 5.5.2.1

Factor $y' x$ out of $y' x$ .

$y' x \cdot 1 + y' x y e^{x - y} = x y e^{x - y} - y$

Step 5.5.2.2

Factor $y' x$ out of $y' x y e^{x - y}$ .

$y' x \cdot 1 + y' x (y e^{x - y}) = x y e^{x - y} - y$

Step 5.5.2.3

Factor $y' x$ out of $y' x \cdot 1 + y' x (y e^{x - y})$ .

$y' x (1 + y e^{x - y}) = x y e^{x - y} - y$

Step 5.5.3

Divide each term in $y' x (1 + y e^{x - y}) = x y e^{x - y} - y$ by $x (1 + y e^{x - y})$ and simplify.

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Step 5.5.3.1

Divide each term in $y' x (1 + y e^{x - y}) = x y e^{x - y} - y$ by $x (1 + y e^{x - y})$ .

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

Step 5.5.3.2

Simplify the left side.

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Step 5.5.3.2.1

Cancel the common factor of $x$ .

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Step 5.5.3.2.1.1

Cancel the common factor.

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

Step 5.5.3.2.1.2

Rewrite the expression.

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

Step 5.5.3.2.2

Cancel the common factor of $1 + y e^{x - y}$ .

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Step 5.5.3.2.2.1

Cancel the common factor.

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

Step 5.5.3.2.2.2

Divide $y'$ by $1$ .

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

Step 5.5.3.3

Simplify the right side.

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Step 5.5.3.3.1

Combine the numerators over the common denominator.

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

Step 5.5.3.3.2

Factor $y$ out of $x y e^{x - y} - y$ .

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Step 5.5.3.3.2.1

Factor $y$ out of $x y e^{x - y}$ .

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

Step 5.5.3.3.2.2

Factor $y$ out of $- y$ .

$y' = \frac{y (x e^{x - y}) + y \cdot - 1}{x (1 + y e^{x - y})}$

Step 5.5.3.3.2.3

Factor $y$ out of $y (x e^{x - y}) + y \cdot - 1$ .

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

Step 6

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

$\frac{d y}{d x} = \frac{y (x e^{x - y} - 1)}{x (1 + y e^{x - y})}$