Find dy/dx x^2y-e^y-4=0

$x^{2} y - e^{y} - 4 = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{2} y - e^{y} - 4) = \frac{d}{d x} (0)$

Step 2

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $x^{2} y - e^{y} - 4$ with respect to $x$ is $\frac{d}{d x} [x^{2} y] + \frac{d}{d x} [- e^{y}] + \frac{d}{d x} [- 4]$ .

$\frac{d}{d x} [x^{2} y] + \frac{d}{d x} [- e^{y}] + \frac{d}{d x} [- 4]$

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

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

$x^{2} \frac{d}{d x} [y] + y \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- e^{y}] + \frac{d}{d x} [- 4]$

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

$x^{2} y' + y \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- e^{y}] + \frac{d}{d x} [- 4]$

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

$x^{2} y' + y (2 x) + \frac{d}{d x} [- e^{y}] + \frac{d}{d x} [- 4]$

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

$x^{2} y' + 2 y x + \frac{d}{d x} [- e^{y}] + \frac{d}{d x} [- 4]$

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

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

$x^{2} y' + 2 y x - \frac{d}{d x} [e^{y}] + \frac{d}{d x} [- 4]$

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 $y$ .

$x^{2} y' + 2 y x - (\frac{d}{d u} [e^{u}] \frac{d}{d x} [y]) + \frac{d}{d x} [- 4]$

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

$x^{2} y' + 2 y x - (e^{u} \frac{d}{d x} [y]) + \frac{d}{d x} [- 4]$

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

$x^{2} y' + 2 y x - (e^{y} \frac{d}{d x} [y]) + \frac{d}{d x} [- 4]$

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

$x^{2} y' + 2 y x - e^{y} y' + \frac{d}{d x} [- 4]$

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

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

Simplify.

Add $x^{2} y' + 2 y x - e^{y} y'$ and $0$ .

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

Reorder terms.

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

Step 3

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

$0$

Step 4

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

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

Step 5

Solve for $y'$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $y'$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 6

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

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

Find dy/dx x^2y-e^y-4=0

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