Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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 $g (x) = y$ .

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.2.4

Multiply $y$ by $1$ .

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

Step 3.3

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

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Step 3.3.1

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 3.3.1.1

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

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

Step 3.3.1.2

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

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

Step 3.3.1.3

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

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

Step 3.3.2

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

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

Step 3.4

Reorder terms.

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

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Since $y'$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

Step 5.2

Subtract $y'$ from both sides of the equation.

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

Step 5.3

Subtract $y$ from both sides of the equation.

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

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

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

Step 5.4.3

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

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

Step 5.4.4

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

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

Step 5.4.5

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

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

Step 5.5

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

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Step 5.5.1

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

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

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

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

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Step 5.5.2.1.1

Cancel the common factor.

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

Step 5.5.2.1.2

Divide $y'$ by $1$ .

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

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Move the negative in front of the fraction.

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

Step 6

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

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