Find dy/dx 9xy+y^2=2x+y

$9 x y + y^{2} = 2 x + y$

Differentiate both sides of the equation.

$\frac{d}{d x} (9 x y + y^{2}) = \frac{d}{d x} (2 x + y)$

Differentiate the left side of the equation.

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By the Sum Rule, the derivative of $9 x y + y^{2}$ with respect to $x$ is $\frac{d}{d x} [9 x y] + \frac{d}{d x} [y^{2}]$ .

$\frac{d}{d x} [9 x y] + \frac{d}{d x} [y^{2}]$

Evaluate $\frac{d}{d x} [9 x y]$ .

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Since $9$ is constant with respect to $x$ , the derivative of $9 x y$ with respect to $x$ is $9 \frac{d}{d x} [x y]$ .

$9 \frac{d}{d x} [x y] + \frac{d}{d x} [y^{2}]$

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 $g (x) = y$ .

$9 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x]) + \frac{d}{d x} [y^{2}]$

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

$9 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x]) + \frac{d}{d x} [y^{2}]$

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

$9 (x \frac{d}{d x} [y] + y \cdot 1) + \frac{d}{d x} [y^{2}]$

Multiply $y$ by $1$ .

$9 (x \frac{d}{d x} [y] + y) + \frac{d}{d x} [y^{2}]$

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

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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 $g (x) = y$ .

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

$9 (x \frac{d}{d x} [y] + y) + \frac{d}{d u} [u^{2}] \frac{d}{d x} [y]$

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

$9 (x \frac{d}{d x} [y] + y) + 2 u \frac{d}{d x} [y]$

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

$9 (x \frac{d}{d x} [y] + y) + 2 y \frac{d}{d x} [y]$

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

$9 (x \frac{d}{d x} [y] + y) + 2 y \frac{d}{d x} [y]$

Simplify.

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Apply the distributive property.

$9 (x \frac{d}{d x} [y]) + 9 y + 2 y \frac{d}{d x} [y]$

Reorder terms.

$9 x \frac{d}{d x} [y] + 2 y \frac{d}{d x} [y] + 9 y$

Differentiate the right side of the equation.

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

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

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

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

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

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

$2 \cdot 1 + \frac{d}{d x} [y]$

Multiply $2$ by $1$ .

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

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

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

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

$9 x y' + 2 y y' + 9 y = 2 + y'$

Solve for $y'$ .

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

$9 x y' + 2 y y' + 9 y - y' = 2$

Subtract $9 y$ from both sides of the equation.

$9 x y' + 2 y y' - y' = 2 - 9 y$

Factor $y'$ out of $9 x y' + 2 y y' - y'$ .

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Factor $y'$ out of $9 x y'$ .

$y' (9 x) + 2 y y' - y' = 2 - 9 y$

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

$y' (9 x) + y' (2 y) - y' = 2 - 9 y$

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

$y' (9 x) + y' (2 y) + y' \cdot - 1 = 2 - 9 y$

Factor $y'$ out of $y' (9 x) + y' (2 y)$ .

$y' (9 x + 2 y) + y' \cdot - 1 = 2 - 9 y$

Factor $y'$ out of $y' (9 x + 2 y) + y' \cdot - 1$ .

$y' (9 x + 2 y - 1) = 2 - 9 y$

Divide each term by $9 x + 2 y - 1$ and simplify.

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Divide each term in $y' (9 x + 2 y - 1) = 2 - 9 y$ by $9 x + 2 y - 1$ .

$\frac{y' (9 x + 2 y - 1)}{9 x + 2 y - 1} = \frac{2}{9 x + 2 y - 1} + \frac{- 9 y}{9 x + 2 y - 1}$

Cancel the common factor of $9 x + 2 y - 1$ .

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Cancel the common factor.

$\frac{y' (9 x + 2 y - 1)}{9 x + 2 y - 1} = \frac{2}{9 x + 2 y - 1} + \frac{- 9 y}{9 x + 2 y - 1}$

Divide $y'$ by $1$ .

$y' = \frac{2}{9 x + 2 y - 1} + \frac{- 9 y}{9 x + 2 y - 1}$

Simplify $\frac{2}{9 x + 2 y - 1} + \frac{- 9 y}{9 x + 2 y - 1}$ .

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Move the negative in front of the fraction.

$y' = \frac{2}{9 x + 2 y - 1} - \frac{9 y}{9 x + 2 y - 1}$

Combine the numerators over the common denominator.

$y' = \frac{2 - 9 y}{9 x + 2 y - 1}$

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

$\frac{d y}{d x} = \frac{2 - 9 y}{9 x + 2 y - 1}$

Find dy/dx 9xy+y^2=2x+y

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