Calculus Examples

$x y^{2} + \frac{x^{2}}{y} = 5$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x y^{2} + \frac{x^{2}}{y}) = \frac{d}{d x} (5)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $x y^{2} + \frac{x^{2}}{y}$ with respect to $x$ is $\frac{d}{d x} [x y^{2}] + \frac{d}{d x} [\frac{x^{2}}{y}]$ .

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

Step 2.2

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

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

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

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

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Step 2.2.2.1

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

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

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 $n = 2$ .

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Move $2$ to the left of $x$ .

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

Step 2.2.6

Multiply $y^{2}$ by $1$ .

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

Step 2.3

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

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Step 2.3.1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x^{2}$ and $g (x) = y$ .

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

Move $2$ to the left of $y$ .

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

Step 2.4

Simplify.

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Step 2.4.1

Combine terms.

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Step 2.4.1.1

To write $2 x y y'$ as a fraction with a common denominator, multiply by $\frac{y^{2}}{y^{2}}$ .

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

Step 2.4.1.2

Combine the numerators over the common denominator.

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

Step 2.4.1.3

Multiply $y$ by $y^{2}$ by adding the exponents.

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Step 2.4.1.3.1

Move $y^{2}$ .

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

Step 2.4.1.3.2

Multiply $y^{2}$ by $y$ .

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Step 2.4.1.3.2.1

Raise $y$ to the power of $1$ .

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

Step 2.4.1.3.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.4.1.3.3

Add $2$ and $1$ .

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

Step 2.4.1.4

To write $y^{2}$ as a fraction with a common denominator, multiply by $\frac{y^{2}}{y^{2}}$ .

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

Step 2.4.1.5

Combine the numerators over the common denominator.

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

Step 2.4.1.6

Multiply $y^{2}$ by $y^{2}$ by adding the exponents.

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Step 2.4.1.6.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.4.1.6.2

Add $2$ and $2$ .

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

Step 2.4.2

Reorder terms.

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

Step 3

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$0$

Step 4

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

$\frac{y^{4} + 2 y^{3} x y' + 2 x y - x^{2} y'}{y^{2}} = 0$

Step 5

Solve for $y'$ .

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Step 5.1

Set the numerator equal to zero.

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

Step 5.2

Solve the equation for $y'$ .

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Step 5.2.1

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

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Step 5.2.1.1

Subtract $y^{4}$ from both sides of the equation.

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

Step 5.2.1.2

Subtract $2 x y$ from both sides of the equation.

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

Step 5.2.2

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

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Step 5.2.2.1

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

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

Step 5.2.2.2

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

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

Step 5.2.2.3

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

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

Step 5.2.3

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

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Step 5.2.3.1

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

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

Step 5.2.3.2

Simplify the left side.

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Step 5.2.3.2.1

Cancel the common factor of $x$ .

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Step 5.2.3.2.1.1

Cancel the common factor.

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

Step 5.2.3.2.1.2

Rewrite the expression.

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

Step 5.2.3.2.2

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

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Step 5.2.3.2.2.1

Cancel the common factor.

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

Step 5.2.3.2.2.2

Divide $y'$ by $1$ .

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

Step 5.2.3.3

Simplify the right side.

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Step 5.2.3.3.1

Simplify each term.

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Step 5.2.3.3.1.1

Move the negative in front of the fraction.

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

Step 5.2.3.3.1.2

Cancel the common factor of $x$ .

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Step 5.2.3.3.1.2.1

Cancel the common factor.

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

Step 5.2.3.3.1.2.2

Rewrite the expression.

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

Step 5.2.3.3.1.3

Move the negative in front of the fraction.

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

Step 5.2.3.3.2

To write $- \frac{2 y}{2 y^{3} - x}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

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

Step 5.2.3.3.3

Write each expression with a common denominator of $x (2 y^{3} - x)$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.2.3.3.3.1

Multiply $\frac{2 y}{2 y^{3} - x}$ by $\frac{x}{x}$ .

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

Step 5.2.3.3.3.2

Reorder the factors of $(2 y^{3} - x) x$ .

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

Step 5.2.3.3.4

Combine the numerators over the common denominator.

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

Step 5.2.3.3.5

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

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Step 5.2.3.3.5.1

Factor $y$ out of $- y^{4}$ .

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

Step 5.2.3.3.5.2

Factor $y$ out of $- 2 y x$ .

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

Step 5.2.3.3.5.3

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

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

Step 5.2.3.3.6

Factor $- 1$ out of $- y^{3}$ .

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

Step 5.2.3.3.7

Factor $- 1$ out of $- 2 x$ .

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

Step 5.2.3.3.8

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

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

Step 5.2.3.3.9

Simplify the expression.

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Step 5.2.3.3.9.1

Rewrite $- (y^{3} + 2 x)$ as $- 1 (y^{3} + 2 x)$ .

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

Step 5.2.3.3.9.2

Move the negative in front of the fraction.

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

Step 6

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

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