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

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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

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

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

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 $g (x) = e^{2 y}$ .

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

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 $g (x) = 2 y$ .

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

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

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

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

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

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

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

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

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

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

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

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

Multiply $e^{2 y}$ by $1$ .

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

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

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

Simplify.

Reorder terms.

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

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

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

Step 3

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

$0$

Step 4

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

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

Step 5

Solve for $y'$ .

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

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

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

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

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

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

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

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

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

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

$y' (2 x e^{2 y}) + y' \cdot 1 = - e^{2 y}$

Factor $y'$ out of $y' (2 x e^{2 y}) + y' \cdot 1$ .

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

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

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

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

Simplify the left side.

Cancel the common factor of $2 x e^{2 y} + 1$ .

Cancel the common factor.

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

Divide $y'$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 6

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

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

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

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