Calculus Examples

$x^{2} y^{3} - 2 x y^{2} = 5$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{2} y^{3} - 2 x y^{2}) = \frac{d}{d x} (5)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $x^{2} y^{3} - 2 x y^{2}$ with respect to $x$ is $\frac{d}{d x} [x^{2} y^{3}] + \frac{d}{d x} [- 2 x y^{2}]$ .

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

Step 2.2

Evaluate $\frac{d}{d x} [x^{2} y^{3}]$ .

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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 $f (x) = x^{2}$ and $g (x) = y^{3}$ .

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

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 $u_{1}$ as $y$ .

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Move $3$ to the left of $x^{2}$ .

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

Step 2.2.6

Move $2$ to the left of $y^{3}$ .

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

Step 2.3

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

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Step 2.3.1

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

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

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

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

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

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Step 2.3.3.1

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

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

Step 2.3.3.2

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

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

Step 2.3.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

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

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

Step 2.3.6

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

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

Step 2.3.7

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

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

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

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

Step 2.4.2

Multiply $2$ by $- 2$ .

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

Step 2.4.3

Reorder terms.

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

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.

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

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

Add $2 y^{2}$ to both sides of the equation.

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

Step 5.1.2

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

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

Factor $x y y'$ out of $- 4 x y y'$ .

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

Step 5.2.3

Factor $x y y'$ out of $x y y' (3 x y) + x y y' \cdot - 4$ .

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

Step 5.3

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

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Step 5.3.1

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

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

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.

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

Step 5.3.2.1.2

Rewrite the expression.

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

Step 5.3.2.2

Cancel the common factor of $y$ .

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Step 5.3.2.2.1

Cancel the common factor.

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

Step 5.3.2.2.2

Rewrite the expression.

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

Step 5.3.2.3

Cancel the common factor of $3 x y - 4$ .

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Step 5.3.2.3.1

Cancel the common factor.

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

Step 5.3.2.3.2

Divide $y'$ by $1$ .

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

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

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

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Step 5.3.3.1.1.1

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

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

Step 5.3.3.1.1.2

Cancel the common factors.

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Step 5.3.3.1.1.2.1

Factor $y$ out of $x y (3 x y - 4)$ .

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

Step 5.3.3.1.1.2.2

Cancel the common factor.

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

Step 5.3.3.1.1.2.3

Rewrite the expression.

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

Step 5.3.3.1.2

Cancel the common factor of $y^{3}$ and $y$ .

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Step 5.3.3.1.2.1

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

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

Step 5.3.3.1.2.2

Cancel the common factors.

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Step 5.3.3.1.2.2.1

Factor $y$ out of $x y (3 x y - 4)$ .

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

Step 5.3.3.1.2.2.2

Cancel the common factor.

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

Step 5.3.3.1.2.2.3

Rewrite the expression.

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

Step 5.3.3.1.3

Cancel the common factor of $x$ .

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Step 5.3.3.1.3.1

Cancel the common factor.

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

Step 5.3.3.1.3.2

Rewrite the expression.

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

Step 5.3.3.1.4

Move the negative in front of the fraction.

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

Step 5.3.3.2

To write $- \frac{2 y^{2}}{3 x y - 4}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

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

Step 5.3.3.3

Write each expression with a common denominator of $x (3 x y - 4)$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.3.3.3.1

Multiply $\frac{2 y^{2}}{3 x y - 4}$ by $\frac{x}{x}$ .

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

Step 5.3.3.3.2

Reorder the factors of $(3 x y - 4) x$ .

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

Step 5.3.3.4

Combine the numerators over the common denominator.

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

Step 5.3.3.5

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

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Step 5.3.3.5.1

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

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

Step 5.3.3.5.2

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

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

Step 5.3.3.5.3

Factor $2 y$ out of $2 y (1) + 2 y (- y x)$ .

$y' = \frac{2 y (1 - y x)}{x (3 x y - 4)}$

Step 6

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

$\frac{d y}{d x} = \frac{2 y (1 - y x)}{x (3 x y - 4)}$