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

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

Step 1

Differentiate both sides of the equation.

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

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

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

$0$

Step 4

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

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

Step 5

Solve for $y'$ .

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

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

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Cancel the common factor.

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

Divide $y'$ by $1$ .

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

Simplify the right side.

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

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

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Step 6

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

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

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

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