Find dx/dy x^3+y^3=7

$x^{3} + y^{3} = 7$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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

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

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

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

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

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

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

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

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

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

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

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

Differentiate using the Power Rule.

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

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

Reorder terms.

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

Step 3

Since $7$ is constant with respect to $y$ , the derivative of $7$ with respect to $y$ is $0$ .

$0$

Step 4

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

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

Step 5

Solve for $x'$ .

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

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

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Cancel the common factor.

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

Divide $x'$ by $1$ .

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

Simplify the right side.

Cancel the common factor of $- 3$ and $3$ .

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

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

Cancel the common factors.

Factor $3$ out of $3 x^{2}$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Step 6

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

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

Find dx/dy x^3+y^3=7

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