Find dy/dx xe^y-ye^x=7

$x e^{y} - y e^{x} = 7$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x e^{y} - y e^{x}) = \frac{d}{d x} (7)$

Step 2

Differentiate the left side of the equation.

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

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

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

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

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

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 $g (x) = y$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Simplify.

Apply the distributive property.

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

Remove unnecessary parentheses.

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

Reorder terms.

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

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 3

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$0$

Step 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 5

Solve for $y'$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Rewrite $- 1 e^{x}$ as $- e^{x}$ .

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $y'$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

Step 6

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

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

Find dy/dx xe^y-ye^x=7

Related Questions

Support LCMGCF.com by disabling your adblocker.