Find dx/dy 4x^3-y^3-5=0

$4 x^{3} - y^{3} - 5 = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (4 x^{3} - y^{3} - 5) = \frac{d}{d y} (0)$

Step 2

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $4 x^{3} - y^{3} - 5$ with respect to $y$ is $\frac{d}{d y} [4 x^{3}] + \frac{d}{d y} [- y^{3}] + \frac{d}{d y} [- 5]$ .

$\frac{d}{d y} [4 x^{3}] + \frac{d}{d y} [- y^{3}] + \frac{d}{d y} [- 5]$

Evaluate $\frac{d}{d y} [4 x^{3}]$ .

Since $4$ is constant with respect to $y$ , the derivative of $4 x^{3}$ with respect to $y$ is $4 \frac{d}{d y} [x^{3}]$ .

$4 \frac{d}{d y} [x^{3}] + \frac{d}{d y} [- y^{3}] + \frac{d}{d y} [- 5]$

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 $x$ .

$4 (\frac{d}{d u} [u^{3}] \frac{d}{d y} [x]) + \frac{d}{d y} [- y^{3}] + \frac{d}{d y} [- 5]$

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

$4 (3 u^{2} \frac{d}{d y} [x]) + \frac{d}{d y} [- y^{3}] + \frac{d}{d y} [- 5]$

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

$4 (3 x^{2} \frac{d}{d y} [x]) + \frac{d}{d y} [- y^{3}] + \frac{d}{d y} [- 5]$

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

$4 (3 x^{2} x') + \frac{d}{d y} [- y^{3}] + \frac{d}{d y} [- 5]$

Multiply $3$ by $4$ .

$12 x^{2} x' + \frac{d}{d y} [- y^{3}] + \frac{d}{d y} [- 5]$

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

Since $- 1$ is constant with respect to $y$ , the derivative of $- y^{3}$ with respect to $y$ is $- \frac{d}{d y} [y^{3}]$ .

$12 x^{2} x' - \frac{d}{d y} [y^{3}] + \frac{d}{d y} [- 5]$

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

$12 x^{2} x' - (3 y^{2}) + \frac{d}{d y} [- 5]$

Multiply $3$ by $- 1$ .

$12 x^{2} x' - 3 y^{2} + \frac{d}{d y} [- 5]$

Since $- 5$ is constant with respect to $y$ , the derivative of $- 5$ with respect to $y$ is $0$ .

$12 x^{2} x' - 3 y^{2} + 0$

Simplify.

Add $12 x^{2} x' - 3 y^{2}$ and $0$ .

$12 x^{2} x' - 3 y^{2}$

Reorder terms.

$- 3 y^{2} + 12 x^{2} x'$

Step 3

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

$0$

Step 4

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

$- 3 y^{2} + 12 x^{2} x' = 0$

Step 5

Solve for $x'$ .

Add $3 y^{2}$ to both sides of the equation.

$12 x^{2} x' = 3 y^{2}$

Divide each term in $12 x^{2} x' = 3 y^{2}$ by $12 x^{2}$ and simplify.

Divide each term in $12 x^{2} x' = 3 y^{2}$ by $12 x^{2}$ .

$\frac{12 x^{2} x'}{12 x^{2}} = \frac{3 y^{2}}{12 x^{2}}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 x^{2} x'}{12 x^{2}} = \frac{3 y^{2}}{12 x^{2}}$

Rewrite the expression.

$\frac{x^{2} x'}{x^{2}} = \frac{3 y^{2}}{12 x^{2}}$

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

Cancel the common factor.

$\frac{x^{2} x'}{x^{2}} = \frac{3 y^{2}}{12 x^{2}}$

Divide $x'$ by $1$ .

$x' = \frac{3 y^{2}}{12 x^{2}}$

Simplify the right side.

Cancel the common factor of $3$ and $12$ .

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

$x' = \frac{3 (y^{2})}{12 x^{2}}$

Cancel the common factors.

Factor $3$ out of $12 x^{2}$ .

$x' = \frac{3 (y^{2})}{3 (4 x^{2})}$

Cancel the common factor.

$x' = \frac{3 y^{2}}{3 (4 x^{2})}$

Rewrite the expression.

$x' = \frac{y^{2}}{4 x^{2}}$

Step 6

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

$\frac{d x}{d y} = \frac{y^{2}}{4 x^{2}}$

Find dx/dy 4x^3-y^3-5=0

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