Find dx/dy y^2-xy-10=0

$y^{2} - x y - 10 = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (y^{2} - x y - 10) = \frac{d}{d y} (0)$

Step 2

Differentiate the left side of the equation.

Differentiate.

By the Sum Rule, the derivative of $y^{2} - x y - 10$ with respect to $y$ is $\frac{d}{d y} [y^{2}] + \frac{d}{d y} [- x y] + \frac{d}{d y} [- 10]$ .

$\frac{d}{d y} [y^{2}] + \frac{d}{d y} [- x y] + \frac{d}{d y} [- 10]$

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

$2 y + \frac{d}{d y} [- x y] + \frac{d}{d y} [- 10]$

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

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

$2 y - \frac{d}{d y} [x y] + \frac{d}{d y} [- 10]$

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 $g (y) = y$ .

$2 y - (x \frac{d}{d y} [y] + y \frac{d}{d y} [x]) + \frac{d}{d y} [- 10]$

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

$2 y - (x \cdot 1 + y \frac{d}{d y} [x]) + \frac{d}{d y} [- 10]$

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

$2 y - (x \cdot 1 + y x') + \frac{d}{d y} [- 10]$

Multiply $x$ by $1$ .

$2 y - (x + y x') + \frac{d}{d y} [- 10]$

Since $- 10$ is constant with respect to $y$ , the derivative of $- 10$ with respect to $y$ is $0$ .

$2 y - (x + y x') + 0$

Simplify.

Apply the distributive property.

$2 y - x - (y x') + 0$

Add $2 y - x - y x'$ and $0$ .

$2 y - x - y x'$

Reorder terms.

$- y x' - x + 2 y$

Step 3

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

$0$

Step 4

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

$- y x' - x + 2 y = 0$

Step 5

Solve for $x'$ .

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

Add $x$ to both sides of the equation.

$- y x' + 2 y = x$

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

$- y x' = x - 2 y$

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

Divide each term in $- y x' = x - 2 y$ by $- 1 y$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Cancel the common factor of $y$ .

Cancel the common factor.

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

Divide $x'$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

Cancel the common factor of $y$ .

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative one from the denominator of $\frac{- 2}{- 1}$ .

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

Rewrite $- 1 \cdot - 2$ as $- - 2$ .

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

Multiply $- 1$ by $- 2$ .

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

Step 6

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

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

Find dx/dy y^2-xy-10=0

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