Calculus Examples

$0 = - 2 y^{2} + x^{2} + x$

Step 1

Differentiate both sides of the equation.

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

Step 2

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

$0$

Step 3

Differentiate the right side of the equation.

Tap for more steps...

Step 3.1

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

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

Step 3.2

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

Tap for more steps...

Step 3.2.1

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

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

Step 3.2.2

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 $g (x) = y$ .

Tap for more steps...

Step 3.2.2.1

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

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

Step 3.2.2.2

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

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

Step 3.2.2.3

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

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

Step 3.2.3

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

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

Step 3.2.4

Multiply $2$ by $- 2$ .

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

Step 3.3

Differentiate using the Power Rule.

Tap for more steps...

Step 3.3.1

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

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

Step 3.3.2

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

$- 4 y y' + 2 x + 1$

Step 4

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

$0 = - 4 y y' + 2 x + 1$

Step 5

Solve for $y'$ .

Tap for more steps...

Step 5.1

Rewrite the equation as $- 4 y y' + 2 x + 1 = 0$ .

$- 4 y y' + 2 x + 1 = 0$

Step 5.2

Move all terms not containing $y'$ to the right side of the equation.

Tap for more steps...

Step 5.2.1

Subtract $2 x$ from both sides of the equation.

$- 4 y y' + 1 = - 2 x$

Step 5.2.2

Subtract $1$ from both sides of the equation.

$- 4 y y' = - 2 x - 1$

Step 5.3

Divide each term in $- 4 y y' = - 2 x - 1$ by $- 4 y$ and simplify.

Tap for more steps...

Step 5.3.1

Divide each term in $- 4 y y' = - 2 x - 1$ by $- 4 y$ .

$\frac{- 4 y y'}{- 4 y} = \frac{- 2 x}{- 4 y} + \frac{- 1}{- 4 y}$

Step 5.3.2

Simplify the left side.

Tap for more steps...

Step 5.3.2.1

Cancel the common factor of $- 4$ .

Tap for more steps...

Step 5.3.2.1.1

Cancel the common factor.

$\frac{- 4 y y'}{- 4 y} = \frac{- 2 x}{- 4 y} + \frac{- 1}{- 4 y}$

Step 5.3.2.1.2

Rewrite the expression.

$\frac{y y'}{y} = \frac{- 2 x}{- 4 y} + \frac{- 1}{- 4 y}$

Step 5.3.2.2

Cancel the common factor of $y$ .

Tap for more steps...

Step 5.3.2.2.1

Cancel the common factor.

$\frac{y y'}{y} = \frac{- 2 x}{- 4 y} + \frac{- 1}{- 4 y}$

Step 5.3.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 2 x}{- 4 y} + \frac{- 1}{- 4 y}$

Step 5.3.3

Simplify the right side.

Tap for more steps...

Step 5.3.3.1

Simplify each term.

Tap for more steps...

Step 5.3.3.1.1

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

Tap for more steps...

Step 5.3.3.1.1.1

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

$y' = \frac{- 2 (x)}{- 4 y} + \frac{- 1}{- 4 y}$

Step 5.3.3.1.1.2

Cancel the common factors.

Tap for more steps...

Step 5.3.3.1.1.2.1

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

$y' = \frac{- 2 (x)}{- 2 (2 y)} + \frac{- 1}{- 4 y}$

Step 5.3.3.1.1.2.2

Cancel the common factor.

$y' = \frac{- 2 x}{- 2 (2 y)} + \frac{- 1}{- 4 y}$

Step 5.3.3.1.1.2.3

Rewrite the expression.

$y' = \frac{x}{2 y} + \frac{- 1}{- 4 y}$

Step 5.3.3.1.2

Dividing two negative values results in a positive value.

$y' = \frac{x}{2 y} + \frac{1}{4 y}$

Step 6

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

$\frac{d y}{d x} = \frac{x}{2 y} + \frac{1}{4 y}$