Calculus Examples

$- 2 x^{2} + 1 + x y^{2} = - 2 y^{3}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (- 2 x^{2} + 1 + x y^{2}) = \frac{d}{d x} (- 2 y^{3})$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $- 2 x^{2} + 1 + x y^{2}$ with respect to $x$ is $\frac{d}{d x} [- 2 x^{2}] + \frac{d}{d x} [1] + \frac{d}{d x} [x y^{2}]$ .

$\frac{d}{d x} [- 2 x^{2}] + \frac{d}{d x} [1] + \frac{d}{d x} [x y^{2}]$

Step 2.2

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

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Step 2.2.1

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

$- 2 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [1] + \frac{d}{d x} [x y^{2}]$

Step 2.2.2

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

$- 2 (2 x) + \frac{d}{d x} [1] + \frac{d}{d x} [x y^{2}]$

Step 2.2.3

Multiply $2$ by $- 2$ .

$- 4 x + \frac{d}{d x} [1] + \frac{d}{d x} [x y^{2}]$

Step 2.3

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$- 4 x + 0 + \frac{d}{d x} [x y^{2}]$

Step 2.4

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

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Step 2.4.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = y^{2}$ .

$- 4 x + 0 + x \frac{d}{d x} [y^{2}] + y^{2} \frac{d}{d x} [x]$

Step 2.4.2

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 $g (x) = y$ .

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Step 2.4.2.1

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

$- 4 x + 0 + x (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x]$

Step 2.4.2.2

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

$- 4 x + 0 + x (2 u \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x]$

Step 2.4.2.3

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

$- 4 x + 0 + x (2 y \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x]$

Step 2.4.3

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

$- 4 x + 0 + x (2 y y') + y^{2} \frac{d}{d x} [x]$

Step 2.4.4

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

$- 4 x + 0 + x (2 y y') + y^{2} \cdot 1$

Step 2.4.5

Move $2$ to the left of $x$ .

$- 4 x + 0 + 2 \cdot x y y' + y^{2} \cdot 1$

Step 2.4.6

Multiply $y^{2}$ by $1$ .

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

Step 2.5

Simplify.

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Step 2.5.1

Add $- 4 x$ and $0$ .

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

Step 2.5.2

Reorder terms.

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 y^{3}$ with respect to $x$ is $- 2 \frac{d}{d x} [y^{3}]$ .

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

Step 3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = y$ .

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.3

Multiply $3$ by $- 2$ .

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

Step 3.4

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

$- 6 y^{2} y'$

Step 4

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

$y^{2} + 2 x y y' - 4 x = - 6 y^{2} y'$

Step 5

Solve for $y'$ .

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Step 5.1

Add $6 y^{2} y'$ to both sides of the equation.

$y^{2} + 2 x y y' - 4 x + 6 y^{2} y' = 0$

Step 5.2

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

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Step 5.2.1

Subtract $y^{2}$ from both sides of the equation.

$2 x y y' - 4 x + 6 y^{2} y' = - y^{2}$

Step 5.2.2

Add $4 x$ to both sides of the equation.

$2 x y y' + 6 y^{2} y' = - y^{2} + 4 x$

Step 5.3

Factor $2 y y'$ out of $2 x y y' + 6 y^{2} y'$ .

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Step 5.3.1

Factor $2 y y'$ out of $2 x y y'$ .

$2 y y' x + 6 y^{2} y' = - y^{2} + 4 x$

Step 5.3.2

Factor $2 y y'$ out of $6 y^{2} y'$ .

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

Step 5.3.3

Factor $2 y y'$ out of $2 y y' x + 2 y y' (3 y)$ .

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $2$ .

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Rewrite the expression.

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

Step 5.4.2.2

Cancel the common factor of $y$ .

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Step 5.4.2.2.1

Cancel the common factor.

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

Step 5.4.2.2.2

Rewrite the expression.

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

Step 5.4.2.3

Cancel the common factor of $x + 3 y$ .

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Step 5.4.2.3.1

Cancel the common factor.

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

Step 5.4.2.3.2

Divide $y'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Simplify each term.

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Step 5.4.3.1.1

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

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Step 5.4.3.1.1.1

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

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

Step 5.4.3.1.1.2

Cancel the common factors.

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Step 5.4.3.1.1.2.1

Factor $y$ out of $2 y (x + 3 y)$ .

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

Step 5.4.3.1.1.2.2

Cancel the common factor.

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

Step 5.4.3.1.1.2.3

Rewrite the expression.

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

Step 5.4.3.1.2

Move the negative in front of the fraction.

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

Step 5.4.3.1.3

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

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Step 5.4.3.1.3.1

Factor $2$ out of $4 x$ .

$y' = - \frac{y}{2 (x + 3 y)} + \frac{2 (2 x)}{2 y (x + 3 y)}$

Step 5.4.3.1.3.2

Cancel the common factors.

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Step 5.4.3.1.3.2.1

Factor $2$ out of $2 y (x + 3 y)$ .

$y' = - \frac{y}{2 (x + 3 y)} + \frac{2 (2 x)}{2 (y (x + 3 y))}$

Step 5.4.3.1.3.2.2

Cancel the common factor.

$y' = - \frac{y}{2 (x + 3 y)} + \frac{2 (2 x)}{2 (y (x + 3 y))}$

Step 5.4.3.1.3.2.3

Rewrite the expression.

$y' = - \frac{y}{2 (x + 3 y)} + \frac{2 x}{y (x + 3 y)}$

Step 6

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

$\frac{d y}{d x} = - \frac{y}{2 (x + 3 y)} + \frac{2 x}{y (x + 3 y)}$