Calculus Examples

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

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x - 4 y + x^{3} + 3 y^{3}) = \frac{d}{d x} (0)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

Differentiate.

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Step 2.1.1

By the Sum Rule, the derivative of $x - 4 y + x^{3} + 3 y^{3}$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 y^{3}]$ .

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

Step 2.1.2

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

$1 + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 y^{3}]$

Step 2.2

Evaluate $\frac{d}{d x} [- 4 y]$ .

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Step 2.2.1

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

$1 - 4 \frac{d}{d x} [y] + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 y^{3}]$

Step 2.2.2

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

$1 - 4 y' + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [3 y^{3}]$

Step 2.3

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

$1 - 4 y' + 3 x^{2} + \frac{d}{d x} [3 y^{3}]$

Step 2.4

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

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Step 2.4.1

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

$1 - 4 y' + 3 x^{2} + 3 \frac{d}{d x} [y^{3}]$

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 $f (x) = x^{3}$ and $g (x) = y$ .

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Step 2.4.2.1

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

$1 - 4 y' + 3 x^{2} + 3 (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y])$

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 $n = 3$ .

$1 - 4 y' + 3 x^{2} + 3 (3 u^{2} \frac{d}{d x} [y])$

Step 2.4.2.3

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

$1 - 4 y' + 3 x^{2} + 3 (3 y^{2} \frac{d}{d x} [y])$

Step 2.4.3

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

$1 - 4 y' + 3 x^{2} + 3 (3 y^{2} y')$

Step 2.4.4

Multiply $3$ by $3$ .

$1 - 4 y' + 3 x^{2} + 9 y^{2} y'$

Step 2.5

Reorder terms.

$3 x^{2} + 9 y^{2} y' + 1 - 4 y'$

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.

$3 x^{2} + 9 y^{2} y' + 1 - 4 y' = 0$

Step 5

Solve for $y'$ .

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Step 5.1

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

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Step 5.1.1

Subtract $3 x^{2}$ from both sides of the equation.

$9 y^{2} y' + 1 - 4 y' = - 3 x^{2}$

Step 5.1.2

Subtract $1$ from both sides of the equation.

$9 y^{2} y' - 4 y' = - 3 x^{2} - 1$

Step 5.2

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

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Step 5.2.1

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

$y' (9 y^{2}) - 4 y' = - 3 x^{2} - 1$

Step 5.2.2

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

$y' (9 y^{2}) + y' \cdot - 4 = - 3 x^{2} - 1$

Step 5.2.3

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

$y' (9 y^{2} - 4) = - 3 x^{2} - 1$

Step 5.3

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

$y' ({(3 y)}^{2} - 4) = - 3 x^{2} - 1$

Step 5.4

Rewrite $4$ as $2^{2}$ .

$y' ({(3 y)}^{2} - 2^{2}) = - 3 x^{2} - 1$

Step 5.5

Factor.

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Step 5.5.1

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

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

Step 5.5.2

Remove unnecessary parentheses.

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

Step 5.6

Divide each term in $y' (3 y + 2) (3 y - 2) = - 3 x^{2} - 1$ by $(3 y + 2) (3 y - 2)$ and simplify.

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Step 5.6.1

Divide each term in $y' (3 y + 2) (3 y - 2) = - 3 x^{2} - 1$ by $(3 y + 2) (3 y - 2)$ .

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

Step 5.6.2

Simplify the left side.

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Step 5.6.2.1

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

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Step 5.6.2.1.1

Cancel the common factor.

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

Step 5.6.2.1.2

Rewrite the expression.

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

Step 5.6.2.2

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

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Step 5.6.2.2.1

Cancel the common factor.

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

Step 5.6.2.2.2

Divide $y'$ by $1$ .

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

Step 5.6.3

Simplify the right side.

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Step 5.6.3.1

Simplify each term.

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Step 5.6.3.1.1

Move the negative in front of the fraction.

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

Step 5.6.3.1.2

Move the negative in front of the fraction.

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

Step 6

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

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