Find dy/dx x^3+y^3=8xy

$x^{3} + y^{3} = 8 x y$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{3} + y^{3}) = \frac{d}{d x} (8 x y)$

Step 2

Differentiate the left side of the equation.

Differentiate.

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

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

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

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

Evaluate $\frac{d}{d x} [y^{3}]$ .

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = y$ .

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

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

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

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

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

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

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

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

Step 3

Differentiate the right side of the equation.

Since $8$ is constant with respect to $x$ , the derivative of $8 x y$ with respect to $x$ is $8 \frac{d}{d x} [x y]$ .

$8 \frac{d}{d x} [x y]$

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 $f (x) = x$ and $g (x) = y$ .

$8 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x])$

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

$8 (x y' + y \frac{d}{d x} [x])$

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

$8 (x y' + y \cdot 1)$

Multiply $y$ by $1$ .

$8 (x y' + y)$

Apply the distributive property.

$8 x y' + 8 y$

Step 4

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

$3 x^{2} + 3 y^{2} y' = 8 x y' + 8 y$

Step 5

Solve for $y'$ .

Subtract $8 x y'$ from both sides of the equation.

$3 x^{2} + 3 y^{2} y' - 8 x y' = 8 y$

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

$3 y^{2} y' - 8 x y' = 8 y - 3 x^{2}$

Factor $y'$ out of $3 y^{2} y' - 8 x y'$ .

Factor $y'$ out of $3 y^{2} y'$ .

$y' (3 y^{2}) - 8 x y' = 8 y - 3 x^{2}$

Factor $y'$ out of $- 8 x y'$ .

$y' (3 y^{2}) + y' (- 8 x) = 8 y - 3 x^{2}$

Factor $y'$ out of $y' (3 y^{2}) + y' (- 8 x)$ .

$y' (3 y^{2} - 8 x) = 8 y - 3 x^{2}$

Divide each term in $y' (3 y^{2} - 8 x) = 8 y - 3 x^{2}$ by $3 y^{2} - 8 x$ and simplify.

Divide each term in $y' (3 y^{2} - 8 x) = 8 y - 3 x^{2}$ by $3 y^{2} - 8 x$ .

$\frac{y' (3 y^{2} - 8 x)}{3 y^{2} - 8 x} = \frac{8 y}{3 y^{2} - 8 x} + \frac{- 3 x^{2}}{3 y^{2} - 8 x}$

Simplify the left side.

Cancel the common factor of $3 y^{2} - 8 x$ .

Cancel the common factor.

$\frac{y' (3 y^{2} - 8 x)}{3 y^{2} - 8 x} = \frac{8 y}{3 y^{2} - 8 x} + \frac{- 3 x^{2}}{3 y^{2} - 8 x}$

Divide $y'$ by $1$ .

$y' = \frac{8 y}{3 y^{2} - 8 x} + \frac{- 3 x^{2}}{3 y^{2} - 8 x}$

Simplify the right side.

Combine the numerators over the common denominator.

$y' = \frac{8 y - 3 x^{2}}{3 y^{2} - 8 x}$

Step 6

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

$\frac{d y}{d x} = \frac{8 y - 3 x^{2}}{3 y^{2} - 8 x}$

Find dy/dx x^3+y^3=8xy

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