Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.3

Differentiate using the Power Rule.

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Step 2.3.1

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

$x^{3} y' + 3 y x^{2} + 3 x^{2}$

Step 2.3.2

Reorder terms.

$3 x^{2} + x^{3} y' + 3 x^{2} y$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- y^{2}$ with respect to $x$ is $- \frac{d}{d x} [y^{2}]$ .

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

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

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.3

Multiply $2$ by $- 1$ .

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

Step 3.4

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

$- 2 y y'$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Add $2 y y'$ to both sides of the equation.

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

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

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

Step 5.2.2

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 5.3.3

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

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

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

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Divide $y'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Combine the numerators over the common denominator.

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

Step 5.4.3.2

Simplify the numerator.

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Step 5.4.3.2.1

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

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Step 5.4.3.2.1.1

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

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

Step 5.4.3.2.1.2

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

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

Step 5.4.3.2.1.3

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

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

Step 5.4.3.2.2

Rewrite $- 1 y$ as $- y$ .

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

Step 5.4.3.3

Simplify with factoring out.

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Step 5.4.3.3.1

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 5.4.3.3.2

Factor $- 1$ out of $- y$ .

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

Step 5.4.3.3.3

Factor $- 1$ out of $- 1 (1) - (y)$ .

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

Step 5.4.3.3.4

Move the negative in front of the fraction.

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

Step 6

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

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