Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

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

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $3$ by $3$ .

$9 x^{2} + \frac{d}{d x} [- x y^{3}] + \frac{d}{d x} [3] + \frac{d}{d x} [- 4 y^{3}]$

Step 2.3

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

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Step 2.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x y^{3}$ with respect to $x$ is $- \frac{d}{d x} [x y^{3}]$ .

$9 x^{2} - \frac{d}{d x} [x y^{3}] + \frac{d}{d x} [3] + \frac{d}{d x} [- 4 y^{3}]$

Step 2.3.2

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

$9 x^{2} - (x \frac{d}{d x} [y^{3}] + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [3] + \frac{d}{d x} [- 4 y^{3}]$

Step 2.3.3

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 2.3.3.1

To apply the Chain Rule, set $u_{1}$ as $y$ .

$9 x^{2} - (x (\frac{d}{d u_{1}} [{u_{1}}^{3}] \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [3] + \frac{d}{d x} [- 4 y^{3}]$

Step 2.3.3.2

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

$9 x^{2} - (x (3 {u_{1}}^{2} \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [3] + \frac{d}{d x} [- 4 y^{3}]$

Step 2.3.3.3

Replace all occurrences of $u_{1}$ with $y$ .

$9 x^{2} - (x (3 y^{2} \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [3] + \frac{d}{d x} [- 4 y^{3}]$

Step 2.3.4

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

$9 x^{2} - (x (3 y^{2} y') + y^{3} \frac{d}{d x} [x]) + \frac{d}{d x} [3] + \frac{d}{d x} [- 4 y^{3}]$

Step 2.3.5

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

$9 x^{2} - (x (3 y^{2} y') + y^{3} \cdot 1) + \frac{d}{d x} [3] + \frac{d}{d x} [- 4 y^{3}]$

Step 2.3.6

Move $3$ to the left of $x$ .

$9 x^{2} - (3 \cdot x y^{2} y' + y^{3} \cdot 1) + \frac{d}{d x} [3] + \frac{d}{d x} [- 4 y^{3}]$

Step 2.3.7

Multiply $y^{3}$ by $1$ .

$9 x^{2} - (3 x y^{2} y' + y^{3}) + \frac{d}{d x} [3] + \frac{d}{d x} [- 4 y^{3}]$

Step 2.4

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$9 x^{2} - (3 x y^{2} y' + y^{3}) + 0 + \frac{d}{d x} [- 4 y^{3}]$

Step 2.5

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

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Step 2.5.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 y^{3}$ with respect to $x$ is $- 4 \frac{d}{d x} [y^{3}]$ .

$9 x^{2} - (3 x y^{2} y' + y^{3}) + 0 - 4 \frac{d}{d x} [y^{3}]$

Step 2.5.2

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 2.5.2.1

To apply the Chain Rule, set $u_{2}$ as $y$ .

$9 x^{2} - (3 x y^{2} y' + y^{3}) + 0 - 4 (\frac{d}{d u_{2}} [{u_{2}}^{3}] \frac{d}{d x} [y])$

Step 2.5.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 3$ .

$9 x^{2} - (3 x y^{2} y' + y^{3}) + 0 - 4 (3 {u_{2}}^{2} \frac{d}{d x} [y])$

Step 2.5.2.3

Replace all occurrences of $u_{2}$ with $y$ .

$9 x^{2} - (3 x y^{2} y' + y^{3}) + 0 - 4 (3 y^{2} \frac{d}{d x} [y])$

Step 2.5.3

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

$9 x^{2} - (3 x y^{2} y' + y^{3}) + 0 - 4 (3 y^{2} y')$

Step 2.5.4

Multiply $3$ by $- 4$ .

$9 x^{2} - (3 x y^{2} y' + y^{3}) + 0 - 12 y^{2} y'$

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

$9 x^{2} - (3 x y^{2} y') - y^{3} + 0 - 12 y^{2} y'$

Step 2.6.2

Combine terms.

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Step 2.6.2.1

Multiply $3$ by $- 1$ .

$9 x^{2} - 3 (x y^{2} y') - y^{3} + 0 - 12 y^{2} y'$

Step 2.6.2.2

Add $9 x^{2} - 3 x y^{2} y' - y^{3}$ and $0$ .

$9 x^{2} - 3 x y^{2} y' - y^{3} - 12 y^{2} y'$

Step 2.6.3

Reorder terms.

$- y^{3} + 9 x^{2} - 3 y^{2} x y' - 12 y^{2} y'$

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.

$- y^{3} + 9 x^{2} - 3 y^{2} x y' - 12 y^{2} y' = 0$

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

Add $y^{3}$ to both sides of the equation.

$9 x^{2} - 3 y^{2} x y' - 12 y^{2} y' = y^{3}$

Step 5.1.2

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

$- 3 y^{2} x y' - 12 y^{2} y' = y^{3} - 9 x^{2}$

Step 5.2

Factor $3 y^{2} y'$ out of $- 3 y^{2} x y' - 12 y^{2} y'$ .

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Step 5.2.1

Factor $3 y^{2} y'$ out of $- 3 y^{2} x y'$ .

$3 y^{2} y' (- 1 x) - 12 y^{2} y' = y^{3} - 9 x^{2}$

Step 5.2.2

Factor $3 y^{2} y'$ out of $- 12 y^{2} y'$ .

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

Step 5.2.3

Factor $3 y^{2} y'$ out of $3 y^{2} y' (- 1 x) + 3 y^{2} y' \cdot - 4$ .

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

Step 5.3

Rewrite $- 1 x$ as $- x$ .

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

Step 5.4

Divide each term in $3 y^{2} y' (- x - 4) = y^{3} - 9 x^{2}$ by $3 y^{2} (- x - 4)$ and simplify.

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Step 5.4.1

Divide each term in $3 y^{2} y' (- x - 4) = y^{3} - 9 x^{2}$ by $3 y^{2} (- x - 4)$ .

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $3$ .

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Rewrite the expression.

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

Step 5.4.2.2

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

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Step 5.4.2.2.1

Cancel the common factor.

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

Step 5.4.2.2.2

Rewrite the expression.

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

Step 5.4.2.3

Cancel the common factor of $- x - 4$ .

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Step 5.4.2.3.1

Cancel the common factor.

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

Step 5.4.2.3.2

Divide $y'$ by $1$ .

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

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 $y^{3}$ and $y^{2}$ .

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Step 5.4.3.1.1.1

Factor $y^{2}$ out of $y^{3}$ .

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

Step 5.4.3.1.1.2

Cancel the common factors.

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Step 5.4.3.1.1.2.1

Factor $y^{2}$ out of $3 y^{2} (- x - 4)$ .

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

Step 5.4.3.1.1.2.2

Cancel the common factor.

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

Step 5.4.3.1.1.2.3

Rewrite the expression.

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

Step 5.4.3.1.2

Cancel the common factor of $- 9$ and $3$ .

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Step 5.4.3.1.2.1

Factor $3$ out of $- 9 x^{2}$ .

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

Step 5.4.3.1.2.2

Cancel the common factors.

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Step 5.4.3.1.2.2.1

Factor $3$ out of $3 y^{2} (- x - 4)$ .

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

Step 5.4.3.1.2.2.2

Cancel the common factor.

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

Step 5.4.3.1.2.2.3

Rewrite the expression.

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

Step 5.4.3.1.3

Move the negative in front of the fraction.

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

Step 6

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

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