Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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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 $f (x) = y$ and $g (x) = e^{x}$ .

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

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

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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 $f (x) = x$ and $g (x) = e^{y}$ .

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

Step 2.3.2

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

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Step 2.3.2.1

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

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

Step 2.3.2.2

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

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

Step 2.3.2.3

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate.

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Step 3.1.1

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

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [\ln (2)]$

Step 3.1.2

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

$2 x + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [\ln (2)]$

Step 3.2

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

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Step 3.2.1

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

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Step 3.2.1.1

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

$2 x + \frac{d}{d u} [u^{2}] \frac{d}{d x} [y] + \frac{d}{d x} [\ln (2)]$

Step 3.2.1.2

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

$2 x + 2 u \frac{d}{d x} [y] + \frac{d}{d x} [\ln (2)]$

Step 3.2.1.3

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

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

Step 3.2.2

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

$2 x + 2 y y' + \frac{d}{d x} [\ln (2)]$

Step 3.3

Since $\ln (2)$ is constant with respect to $x$ , the derivative of $\ln (2)$ with respect to $x$ is $0$ .

$2 x + 2 y y' + 0$

Step 3.4

Simplify.

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Step 3.4.1

Add $2 x + 2 y y'$ and $0$ .

$2 x + 2 y y'$

Step 3.4.2

Reorder terms.

$2 y y' + 2 x$

Step 4

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

$y e^{x} + e^{x} y' + x e^{y} y' + e^{y} = 2 y y' + 2 x$

Step 5

Solve for $y'$ .

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Step 5.1

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

$y e^{x} + y' e^{x} + x y' e^{y} + e^{y} = 2 y y' + 2 x$

Step 5.2

Subtract $2 y y'$ from both sides of the equation.

$y e^{x} + y' e^{x} + x y' e^{y} + e^{y} - 2 y y' = 2 x$

Step 5.3

Move all terms not containing $y'$ to the right side of the equation.

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Step 5.3.1

Subtract $y e^{x}$ from both sides of the equation.

$y' e^{x} + x y' e^{y} + e^{y} - 2 y y' = 2 x - y e^{x}$

Step 5.3.2

Subtract $e^{y}$ from both sides of the equation.

$y' e^{x} + x y' e^{y} - 2 y y' = 2 x - y e^{x} - e^{y}$

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

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

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

Step 5.4.3

Factor $y'$ out of $- 2 y y'$ .

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

Step 5.4.4

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

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

Step 5.4.5

Factor $y'$ out of $y' (e^{x} + x e^{y}) + y' (- 2 y)$ .

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

Step 5.5

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

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Step 5.5.1

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

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

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $e^{x} + x e^{y} - 2 y$ .

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Step 5.5.2.1.1

Cancel the common factor.

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

Step 5.5.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{2 x}{e^{x} + x e^{y} - 2 y} + \frac{- y e^{x}}{e^{x} + x e^{y} - 2 y} + \frac{- e^{y}}{e^{x} + x e^{y} - 2 y}$

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Simplify each term.

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Step 5.5.3.1.1

Move the negative in front of the fraction.

$y' = \frac{2 x}{e^{x} + x e^{y} - 2 y} - \frac{y e^{x}}{e^{x} + x e^{y} - 2 y} + \frac{- e^{y}}{e^{x} + x e^{y} - 2 y}$

Step 5.5.3.1.2

Move the negative in front of the fraction.

$y' = \frac{2 x}{e^{x} + x e^{y} - 2 y} - \frac{y e^{x}}{e^{x} + x e^{y} - 2 y} - \frac{e^{y}}{e^{x} + x e^{y} - 2 y}$

Step 5.5.3.2

Combine into one fraction.

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Step 5.5.3.2.1

Combine the numerators over the common denominator.

$y' = \frac{2 x - y e^{x}}{e^{x} + x e^{y} - 2 y} - \frac{e^{y}}{e^{x} + x e^{y} - 2 y}$

Step 5.5.3.2.2

Combine the numerators over the common denominator.

$y' = \frac{2 x - y e^{x} - e^{y}}{e^{x} + x e^{y} - 2 y}$

Step 6

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

$\frac{d y}{d x} = \frac{2 x - y e^{x} - e^{y}}{e^{x} + x e^{y} - 2 y}$