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

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

Step 1

Differentiate both sides of the equation.

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

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

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

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

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

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

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

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

Multiply $2$ by $2$ .

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

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

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

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

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 $y$ .

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

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

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

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

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

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

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

Multiply $3$ by $- 1$ .

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

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

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

Simplify.

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

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

Reorder terms.

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

Step 3

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

$0$

Step 4

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

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

Step 5

Solve for $y'$ .

Subtract $4 x$ from both sides of the equation.

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

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

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

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

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Cancel the common factor.

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

Divide $y'$ by $1$ .

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

Simplify the right side.

Dividing two negative values results in a positive value.

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

Step 6

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

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

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

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