Find dx/dy xe^(2y)+y=4

$x e^{2 y} + y = 4$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (x e^{2 y} + y) = \frac{d}{d y} (4)$

Step 2

Differentiate the left side of the equation.

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

$\frac{d}{d y} [x e^{2 y}] + \frac{d}{d y} [y]$

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

Differentiate using the Product Rule which states that $\frac{d}{d y} [f (y) g (y)]$ is $f (y) \frac{d}{d y} [g (y)] + g (y) \frac{d}{d y} [f (y)]$ where $f (y) = x$ and $g (y) = e^{2 y}$ .

$x \frac{d}{d y} [e^{2 y}] + e^{2 y} \frac{d}{d y} [x] + \frac{d}{d y} [y]$

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

To apply the Chain Rule, set $u$ as $2 y$ .

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

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

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

Replace all occurrences of $u$ with $2 y$ .

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

Since $2$ is constant with respect to $y$ , the derivative of $2 y$ with respect to $y$ is $2 \frac{d}{d y} [y]$ .

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

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

$x (e^{2 y} (2 \cdot 1)) + e^{2 y} \frac{d}{d y} [x] + \frac{d}{d y} [y]$

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

$x (e^{2 y} (2 \cdot 1)) + e^{2 y} x' + \frac{d}{d y} [y]$

Multiply $2$ by $1$ .

$x (e^{2 y} \cdot 2) + e^{2 y} x' + \frac{d}{d y} [y]$

Move $2$ to the left of $e^{2 y}$ .

$x (2 \cdot e^{2 y}) + e^{2 y} x' + \frac{d}{d y} [y]$

Move $2$ to the left of $x$ .

$2 x e^{2 y} + e^{2 y} x' + \frac{d}{d y} [y]$

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

$2 x e^{2 y} + e^{2 y} x' + 1$

Step 3

Since $4$ is constant with respect to $y$ , the derivative of $4$ with respect to $y$ is $0$ .

$0$

Step 4

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

$2 x e^{2 y} + e^{2 y} x' + 1 = 0$

Step 5

Solve for $x'$ .

Reorder factors in $2 x e^{2 y} + e^{2 y} x' + 1$ .

$2 x e^{2 y} + x' e^{2 y} + 1 = 0$

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

Subtract $2 x e^{2 y}$ from both sides of the equation.

$x' e^{2 y} + 1 = - 2 x e^{2 y}$

Subtract $1$ from both sides of the equation.

$x' e^{2 y} = - 2 x e^{2 y} - 1$

Divide each term in $x' e^{2 y} = - 2 x e^{2 y} - 1$ by $e^{2 y}$ and simplify.

Divide each term in $x' e^{2 y} = - 2 x e^{2 y} - 1$ by $e^{2 y}$ .

$\frac{x' e^{2 y}}{e^{2 y}} = \frac{- 2 x e^{2 y}}{e^{2 y}} + \frac{- 1}{e^{2 y}}$

Simplify the left side.

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

Cancel the common factor.

$\frac{x' e^{2 y}}{e^{2 y}} = \frac{- 2 x e^{2 y}}{e^{2 y}} + \frac{- 1}{e^{2 y}}$

Divide $x'$ by $1$ .

$x' = \frac{- 2 x e^{2 y}}{e^{2 y}} + \frac{- 1}{e^{2 y}}$

Simplify the right side.

Simplify each term.

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

Cancel the common factor.

$x' = \frac{- 2 x e^{2 y}}{e^{2 y}} + \frac{- 1}{e^{2 y}}$

Divide $- 2 x$ by $1$ .

$x' = - 2 x + \frac{- 1}{e^{2 y}}$

Move the negative in front of the fraction.

$x' = - 2 x - \frac{1}{e^{2 y}}$

Step 6

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

$\frac{d x}{d y} = - 2 x - \frac{1}{e^{2 y}}$

Find dx/dy xe^(2y)+y=4

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