Find dx/dy 2x^2-y^3-7=0

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

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (2 x^{2} - y^{3} - 7) = \frac{d}{d y} (0)$

Step 2

Differentiate the left side of the equation.

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

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

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

Since $2$ is constant with respect to $y$ , the derivative of $2 x^{2}$ with respect to $y$ is $2 \frac{d}{d y} [x^{2}]$ .

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

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 $x$ .

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

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

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

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

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

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

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

Multiply $2$ by $2$ .

$4 x x' + \frac{d}{d y} [- y^{3}] + \frac{d}{d y} [- 7]$

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

$4 x x' - \frac{d}{d y} [y^{3}] + \frac{d}{d y} [- 7]$

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

$4 x x' - (3 y^{2}) + \frac{d}{d y} [- 7]$

Multiply $3$ by $- 1$ .

$4 x x' - 3 y^{2} + \frac{d}{d y} [- 7]$

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

$4 x x' - 3 y^{2} + 0$

Simplify.

Add $4 x x' - 3 y^{2}$ and $0$ .

$4 x x' - 3 y^{2}$

Reorder terms.

$- 3 y^{2} + 4 x x'$

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.

$- 3 y^{2} + 4 x x' = 0$

Step 5

Solve for $x'$ .

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

$4 x x' = 3 y^{2}$

Divide each term in $4 x x' = 3 y^{2}$ by $4 x$ and simplify.

Divide each term in $4 x x' = 3 y^{2}$ by $4 x$ .

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x$ .

Cancel the common factor.

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

Divide $x'$ by $1$ .

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

Step 6

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

$\frac{d x}{d y} = \frac{3 y^{2}}{4 x}$

Find dx/dy 2x^2-y^3-7=0

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