Find dy/dx x^2y-y^2=1

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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

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

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 $g (x) = y$ .

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

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

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

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

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

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

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

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

Since $- 1$ is constant with respect to $x$ , the derivative of $- y^{2}$ with respect to $x$ is $- \frac{d}{d x} [y^{2}]$ .

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

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

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

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

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

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

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

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

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

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

Multiply $2$ by $- 1$ .

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

Reorder terms.

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

Step 3

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

$0$

Step 4

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

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

Step 5

Solve for $y'$ .

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

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

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

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

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

Factor $y'$ out of $- 2 y y'$ .

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

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

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $y'$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 6

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

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

Find dy/dx x^2y-y^2=1

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