Calculus Examples

$x e^{y} + y e^{x} = 1$ , $(0, 1)$

Step 1

Find the first derivative and evaluate at $x = 0$ and $y = 1$ to find the slope of the tangent line.

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Step 1.1

Differentiate both sides of the equation.

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

Step 1.2

Differentiate the left side of the equation.

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Step 1.2.1

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

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

Step 1.2.2

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

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Step 1.2.2.1

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

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

Step 1.2.2.2

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

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Step 1.2.2.2.1

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

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

Step 1.2.2.2.2

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

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

Step 1.2.2.2.3

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

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

Step 1.2.2.3

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

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

Step 1.2.2.4

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

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

Step 1.2.2.5

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

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

Step 1.2.3

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

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Step 1.2.3.1

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 $g (x) = e^{x}$ .

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

Step 1.2.3.2

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

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

Step 1.2.3.3

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

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

Step 1.2.4

Simplify.

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Step 1.2.4.1

Reorder terms.

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

Step 1.2.4.2

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

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

Step 1.3

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

$0$

Step 1.4

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

$x e^{y} y' + y e^{x} + e^{x} y' + e^{y} = 0$

Step 1.5

Solve for $y'$ .

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Step 1.5.1

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

$x y' e^{y} + y e^{x} + y' e^{x} + e^{y} = 0$

Step 1.5.2

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

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Step 1.5.2.1

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

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

Step 1.5.2.2

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

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

Step 1.5.3

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

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Step 1.5.3.1

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

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

Step 1.5.3.2

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

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

Step 1.5.3.3

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

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

Step 1.5.4

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

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Step 1.5.4.1

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

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

Step 1.5.4.2

Simplify the left side.

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Step 1.5.4.2.1

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

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Step 1.5.4.2.1.1

Cancel the common factor.

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

Step 1.5.4.2.1.2

Divide $y'$ by $1$ .

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

Step 1.5.4.3

Simplify the right side.

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Step 1.5.4.3.1

Combine the numerators over the common denominator.

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

Step 1.5.4.3.2

Factor $- 1$ out of $- y e^{x}$ .

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

Step 1.5.4.3.3

Factor $- 1$ out of $- e^{y}$ .

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

Step 1.5.4.3.4

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

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

Step 1.5.4.3.5

Simplify the expression.

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Step 1.5.4.3.5.1

Rewrite $- (y e^{x} + e^{y})$ as $- 1 (y e^{x} + e^{y})$ .

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

Step 1.5.4.3.5.2

Move the negative in front of the fraction.

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

Step 1.6

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

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

Step 1.7

Evaluate at $x = 0$ and $y = 1$ .

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Step 1.7.1

Replace the variable $x$ with $0$ in the expression.

$- \frac{y e^{0} + e^{y}}{(0) \cdot e^{y} + e^{0}}$

Step 1.7.2

Replace the variable $y$ with $1$ in the expression.

$- \frac{(1) \cdot e^{0} + e^{1}}{(0) \cdot e^{1} + e^{0}}$

Step 1.7.3

Simplify the numerator.

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Step 1.7.3.1

Multiply $e^{0}$ by $1$ .

$- \frac{e^{0} + e^{1}}{0 \cdot e^{1} + e^{0}}$

Step 1.7.3.2

Anything raised to $0$ is $1$ .

$- \frac{1 + e^{1}}{0 \cdot e^{1} + e^{0}}$

Step 1.7.3.3

Simplify.

$- \frac{1 + e}{0 \cdot e^{1} + e^{0}}$

Step 1.7.4

Simplify the denominator.

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Step 1.7.4.1

Simplify.

$- \frac{1 + e}{0 \cdot e + e^{0}}$

Step 1.7.4.2

Multiply $0$ by $e$ .

$- \frac{1 + e}{0 + e^{0}}$

Step 1.7.4.3

Anything raised to $0$ is $1$ .

$- \frac{1 + e}{0 + 1}$

Step 1.7.4.4

Add $0$ and $1$ .

$- \frac{1 + e}{1}$

Step 1.7.5

Simplify by multiplying through.

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Step 1.7.5.1

Divide $1 + e$ by $1$ .

$- (1 + e)$

Step 1.7.5.2

Apply the distributive property.

$- 1 \cdot 1 - e$

Step 1.7.5.3

Multiply $- 1$ by $1$ .

$- 1 - e$

Step 2

Plug the slope and point values into the point-slope formula and solve for $y$ .

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Step 2.1

Use the slope $- 1 - e$ and a given point $(0, 1)$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (1) = (- 1 - e) \cdot (x - (0))$

Step 2.2

Simplify the equation and keep it in point-slope form.

$y - 1 = (- 1 - e) \cdot (x + 0)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Simplify $(- 1 - e) \cdot (x + 0)$ .

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Step 2.3.1.1

Add $x$ and $0$ .

$y - 1 = (- 1 - e) \cdot x$

Step 2.3.1.2

Apply the distributive property.

$y - 1 = - 1 x - e x$

Step 2.3.1.3

Rewrite $- 1 x$ as $- x$ .

$y - 1 = - x - e x$

Step 2.3.2

Add $1$ to both sides of the equation.

$y = - x - e x + 1$

Step 2.3.3

Write in $y = m x + b$ form.

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Step 2.3.3.1

Move $e$ .

$y = - x - 1 x e + 1$

Step 2.3.3.2

Rewrite $- 1 x$ as $- x$ .

$y = - x - x e + 1$

Step 2.3.3.3

Factor $x$ out of $- x$ .

$y = x \cdot - 1 - x e + 1$

Step 2.3.3.4

Factor $x$ out of $- x e$ .

$y = x \cdot - 1 + x (- e) + 1$

Step 2.3.3.5

Factor $x$ out of $x \cdot - 1 + x (- e)$ .

$y = x \cdot (- 1 - e) + 1$

Step 2.3.3.6

Multiply $- 1 - e$ by $x$ .

$y = (- 1 - e) x + 1$

Step 3