Find dy/dx 2x^2-y^2=6

$2 x^{2} - y^{2} = 6$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (2 x^{2} - y^{2}) = \frac{d}{d x} (6)$

Step 2

Differentiate the left side of the equation.

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

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

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

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

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

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

Multiply $2$ by $2$ .

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

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

Since $- 1$ is constant with respect to $x$ , the derivative of $- y^{2}$ with respect to $x$ is $- \frac{d}{d x} [y^{2}]$ .

$4 x - \frac{d}{d x} [y^{2}]$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = y$ .

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

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

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

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

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

$4 x - (2 y \frac{d}{d x} [y])$

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

$4 x - (2 y y')$

Multiply $2$ by $- 1$ .

$4 x - 2 y y'$

Reorder terms.

$- 2 y y' + 4 x$

Step 3

Since $6$ is constant with respect to $x$ , the derivative of $6$ with respect to $x$ is $0$ .

$0$

Step 4

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

$- 2 y y' + 4 x = 0$

Step 5

Solve for $y'$ .

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

$- 2 y y' = - 4 x$

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

Divide each term in $- 2 y y' = - 4 x$ by $- 2 y$ .

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $y$ .

Cancel the common factor.

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

Divide $y'$ by $1$ .

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

Simplify the right side.

Cancel the common factor of $- 4$ and $- 2$ .

Factor $- 2$ out of $- 4 x$ .

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

Cancel the common factors.

Factor $- 2$ out of $- 2 y$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 6

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

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

Find dy/dx 2x^2-y^2=6

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