Calculus Examples

$x y^{3} + x y = 8$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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

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

Step 2.2.2

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 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.3

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

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Step 2.3.1

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 $g (x) = y$ .

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

Step 2.3.2

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

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

Step 2.3.3

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

$3 x y^{2} y' + y^{3} + x y' + y \cdot 1$

Step 2.3.4

Multiply $y$ by $1$ .

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

Step 2.4

Reorder terms.

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

Step 3

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$0$

Step 4

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

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

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

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

Step 5.1.2

Subtract $y$ from both sides of the equation.

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.2.3

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $x$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Rewrite the expression.

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

Step 5.3.2.2

Cancel the common factor of $3 y^{2} + 1$ .

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Step 5.3.2.2.1

Cancel the common factor.

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

Step 5.3.2.2.2

Divide $y'$ by $1$ .

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Simplify each term.

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Step 5.3.3.1.1

Move the negative in front of the fraction.

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

Step 5.3.3.1.2

Move the negative in front of the fraction.

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

Step 6

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

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