Find dx/dy x^2-7xy+y^3=C

$x^{2} - 7 x y + y^{3} = C$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (x^{2} - 7 x y + y^{3}) = \frac{d}{d y} (C)$

Step 2

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $x^{2} - 7 x y + y^{3}$ with respect to $y$ is $\frac{d}{d y} [x^{2}] + \frac{d}{d y} [- 7 x y] + \frac{d}{d y} [y^{3}]$ .

$\frac{d}{d y} [x^{2}] + \frac{d}{d y} [- 7 x y] + \frac{d}{d y} [y^{3}]$

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

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

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

$\frac{d}{d u} [u^{2}] \frac{d}{d y} [x] + \frac{d}{d y} [- 7 x y] + \frac{d}{d y} [y^{3}]$

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

$2 u \frac{d}{d y} [x] + \frac{d}{d y} [- 7 x y] + \frac{d}{d y} [y^{3}]$

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

$2 x \frac{d}{d y} [x] + \frac{d}{d y} [- 7 x y] + \frac{d}{d y} [y^{3}]$

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

$2 x x' + \frac{d}{d y} [- 7 x y] + \frac{d}{d y} [y^{3}]$

Evaluate $\frac{d}{d y} [- 7 x y]$ .

Since $- 7$ is constant with respect to $y$ , the derivative of $- 7 x y$ with respect to $y$ is $- 7 \frac{d}{d y} [x y]$ .

$2 x x' - 7 \frac{d}{d y} [x y] + \frac{d}{d y} [y^{3}]$

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 $g (y) = y$ .

$2 x x' - 7 (x \frac{d}{d y} [y] + y \frac{d}{d y} [x]) + \frac{d}{d y} [y^{3}]$

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

$2 x x' - 7 (x \cdot 1 + y \frac{d}{d y} [x]) + \frac{d}{d y} [y^{3}]$

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

$2 x x' - 7 (x \cdot 1 + y x') + \frac{d}{d y} [y^{3}]$

Multiply $x$ by $1$ .

$2 x x' - 7 (x + y x') + \frac{d}{d y} [y^{3}]$

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 3$ .

$2 x x' - 7 (x + y x') + 3 y^{2}$

Simplify.

Apply the distributive property.

$2 x x' - 7 x - 7 (y x') + 3 y^{2}$

Remove unnecessary parentheses.

$2 x x' - 7 x - 7 y x' + 3 y^{2}$

Reorder terms.

$3 y^{2} + 2 x x' - 7 y x' - 7 x$

Step 3

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

$0$

Step 4

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

$3 y^{2} + 2 x x' - 7 y x' - 7 x = 0$

Step 5

Solve for $x'$ .

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

Subtract $3 y^{2}$ from both sides of the equation.

$2 x x' - 7 y x' - 7 x = - 3 y^{2}$

Add $7 x$ to both sides of the equation.

$2 x x' - 7 y x' = - 3 y^{2} + 7 x$

Factor $x'$ out of $2 x x' - 7 y x'$ .

Factor $x'$ out of $2 x x'$ .

$x' (2 x) - 7 y x' = - 3 y^{2} + 7 x$

Factor $x'$ out of $- 7 y x'$ .

$x' (2 x) + x' (- 7 y) = - 3 y^{2} + 7 x$

Factor $x'$ out of $x' (2 x) + x' (- 7 y)$ .

$x' (2 x - 7 y) = - 3 y^{2} + 7 x$

Divide each term in $x' (2 x - 7 y) = - 3 y^{2} + 7 x$ by $2 x - 7 y$ and simplify.

Divide each term in $x' (2 x - 7 y) = - 3 y^{2} + 7 x$ by $2 x - 7 y$ .

$\frac{x' (2 x - 7 y)}{2 x - 7 y} = \frac{- 3 y^{2}}{2 x - 7 y} + \frac{7 x}{2 x - 7 y}$

Simplify the left side.

Cancel the common factor of $2 x - 7 y$ .

Cancel the common factor.

$\frac{x' (2 x - 7 y)}{2 x - 7 y} = \frac{- 3 y^{2}}{2 x - 7 y} + \frac{7 x}{2 x - 7 y}$

Divide $x'$ by $1$ .

$x' = \frac{- 3 y^{2}}{2 x - 7 y} + \frac{7 x}{2 x - 7 y}$

Simplify the right side.

Combine the numerators over the common denominator.

$x' = \frac{- 3 y^{2} + 7 x}{2 x - 7 y}$

Factor $- 1$ out of $- 3 y^{2}$ .

$x' = \frac{- (3 y^{2}) + 7 x}{2 x - 7 y}$

Factor $- 1$ out of $7 x$ .

$x' = \frac{- (3 y^{2}) - (- 7 x)}{2 x - 7 y}$

Factor $- 1$ out of $- (3 y^{2}) - (- 7 x)$ .

$x' = \frac{- (3 y^{2} - 7 x)}{2 x - 7 y}$

Simplify the expression.

Rewrite $- (3 y^{2} - 7 x)$ as $- 1 (3 y^{2} - 7 x)$ .

$x' = \frac{- 1 (3 y^{2} - 7 x)}{2 x - 7 y}$

Move the negative in front of the fraction.

$x' = - \frac{3 y^{2} - 7 x}{2 x - 7 y}$

Step 6

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

$\frac{d x}{d y} = - \frac{3 y^{2} - 7 x}{2 x - 7 y}$

Find dx/dy x^2-7xy+y^3=C

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