Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $3$ by $3$ .

$9 x^{2} + \frac{d}{d x} [- 3 y^{3}]$

Step 2.3

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

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Step 2.3.1

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

$9 x^{2} - 3 \frac{d}{d x} [y^{3}]$

Step 2.3.2

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 2.3.2.1

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

$9 x^{2} - 3 (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y])$

Step 2.3.2.2

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

$9 x^{2} - 3 (3 u^{2} \frac{d}{d x} [y])$

Step 2.3.2.3

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

$9 x^{2} - 3 (3 y^{2} \frac{d}{d x} [y])$

Step 2.3.3

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

$9 x^{2} - 3 (3 y^{2} y')$

Step 2.3.4

Multiply $3$ by $- 3$ .

$9 x^{2} - 9 y^{2} y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x^{2} y$ with respect to $x$ is $- 4 \frac{d}{d x} [x^{2} y]$ .

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

Step 3.2

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 $f (x) = x^{2}$ and $g (x) = y$ .

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

Simplify.

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Step 3.6.1

Apply the distributive property.

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

Step 3.6.2

Multiply $2$ by $- 4$ .

$- 4 x^{2} y' - 8 y x$

Step 3.6.3

Reorder terms.

$- 4 x^{2} y' - 8 x y$

Step 4

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

$9 x^{2} - 9 y^{2} y' = - 4 x^{2} y' - 8 x y$

Step 5

Solve for $y'$ .

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Step 5.1

Add $4 x^{2} y'$ to both sides of the equation.

$9 x^{2} - 9 y^{2} y' + 4 x^{2} y' = - 8 x y$

Step 5.2

Subtract $9 x^{2}$ from both sides of the equation.

$- 9 y^{2} y' + 4 x^{2} y' = - 8 x y - 9 x^{2}$

Step 5.3

Factor $y'$ out of $- 9 y^{2} y' + 4 x^{2} y'$ .

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Step 5.3.1

Factor $y'$ out of $- 9 y^{2} y'$ .

$y' (- 9 y^{2}) + 4 x^{2} y' = - 8 x y - 9 x^{2}$

Step 5.3.2

Factor $y'$ out of $4 x^{2} y'$ .

$y' (- 9 y^{2}) + y' (4 x^{2}) = - 8 x y - 9 x^{2}$

Step 5.3.3

Factor $y'$ out of $y' (- 9 y^{2}) + y' (4 x^{2})$ .

$y' (- 9 y^{2} + 4 x^{2}) = - 8 x y - 9 x^{2}$

Step 5.4

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

$y' (- 9 y^{2} + {(2 x)}^{2}) = - 8 x y - 9 x^{2}$

Step 5.5

Rewrite $9 y^{2}$ as ${(3 y)}^{2}$ .

$y' (- {(3 y)}^{2} + {(2 x)}^{2}) = - 8 x y - 9 x^{2}$

Step 5.6

Reorder $- {(3 y)}^{2}$ and ${(2 x)}^{2}$ .

$y' ({(2 x)}^{2} - {(3 y)}^{2}) = - 8 x y - 9 x^{2}$

Step 5.7

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 x$ and $b = 3 y$ .

$y' ((2 x + 3 y) (2 x - (3 y))) = - 8 x y - 9 x^{2}$

Step 5.8

Factor.

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Step 5.8.1

Multiply $3$ by $- 1$ .

$y' ((2 x + 3 y) (2 x - 3 y)) = - 8 x y - 9 x^{2}$

Step 5.8.2

Remove unnecessary parentheses.

$y' (2 x + 3 y) (2 x - 3 y) = - 8 x y - 9 x^{2}$

Step 5.9

Divide each term in $y' (2 x + 3 y) (2 x - 3 y) = - 8 x y - 9 x^{2}$ by $(2 x + 3 y) (2 x - 3 y)$ and simplify.

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Step 5.9.1

Divide each term in $y' (2 x + 3 y) (2 x - 3 y) = - 8 x y - 9 x^{2}$ by $(2 x + 3 y) (2 x - 3 y)$ .

$\frac{y' (2 x + 3 y) (2 x - 3 y)}{(2 x + 3 y) (2 x - 3 y)} = \frac{- 8 x y}{(2 x + 3 y) (2 x - 3 y)} + \frac{- 9 x^{2}}{(2 x + 3 y) (2 x - 3 y)}$

Step 5.9.2

Simplify the left side.

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Step 5.9.2.1

Cancel the common factor of $2 x + 3 y$ .

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Step 5.9.2.1.1

Cancel the common factor.

$\frac{y' (2 x + 3 y) (2 x - 3 y)}{(2 x + 3 y) (2 x - 3 y)} = \frac{- 8 x y}{(2 x + 3 y) (2 x - 3 y)} + \frac{- 9 x^{2}}{(2 x + 3 y) (2 x - 3 y)}$

Step 5.9.2.1.2

Rewrite the expression.

$\frac{y' (2 x - 3 y)}{2 x - 3 y} = \frac{- 8 x y}{(2 x + 3 y) (2 x - 3 y)} + \frac{- 9 x^{2}}{(2 x + 3 y) (2 x - 3 y)}$

Step 5.9.2.2

Cancel the common factor of $2 x - 3 y$ .

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Step 5.9.2.2.1

Cancel the common factor.

$\frac{y' (2 x - 3 y)}{2 x - 3 y} = \frac{- 8 x y}{(2 x + 3 y) (2 x - 3 y)} + \frac{- 9 x^{2}}{(2 x + 3 y) (2 x - 3 y)}$

Step 5.9.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 8 x y}{(2 x + 3 y) (2 x - 3 y)} + \frac{- 9 x^{2}}{(2 x + 3 y) (2 x - 3 y)}$

Step 5.9.3

Simplify the right side.

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Step 5.9.3.1

Simplify each term.

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Step 5.9.3.1.1

Move the negative in front of the fraction.

$y' = - \frac{8 x y}{(2 x + 3 y) (2 x - 3 y)} + \frac{- 9 x^{2}}{(2 x + 3 y) (2 x - 3 y)}$

Step 5.9.3.1.2

Move the negative in front of the fraction.

$y' = - \frac{8 x y}{(2 x + 3 y) (2 x - 3 y)} - \frac{9 x^{2}}{(2 x + 3 y) (2 x - 3 y)}$

Step 6

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

$\frac{d y}{d x} = - \frac{8 x y}{(2 x + 3 y) (2 x - 3 y)} - \frac{9 x^{2}}{(2 x + 3 y) (2 x - 3 y)}$