Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{2}$ and $g (y) = x$ .

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Step 2.2.3.1

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

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

Step 2.2.3.2

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

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

Step 2.2.3.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.4

Reorder terms.

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $3$ is constant with respect to $y$ , the derivative of $3 x$ with respect to $y$ is $3 \frac{d}{d y} [x]$ .

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

Step 3.2

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

$3 x'$

Step 4

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

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

Step 5

Solve for $x'$ .

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Step 5.1

Subtract $3 x'$ from both sides of the equation.

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 5.3.3

Factor $x'$ out of $- 3 x'$ .

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

Step 5.3.4

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

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

Step 5.3.5

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

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

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

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Divide $x'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Combine the numerators over the common denominator.

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

Step 5.4.3.2

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

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Step 5.4.3.2.1

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

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

Step 5.4.3.2.2

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

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

Step 5.4.3.2.3

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

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

Step 5.4.3.3

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

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

Step 5.4.3.4

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

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

Step 5.4.3.5

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

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

Step 5.4.3.6

Simplify the expression.

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Step 5.4.3.6.1

Rewrite $- (x + 2 y)$ as $- 1 (x + 2 y)$ .

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

Step 5.4.3.6.2

Move the negative in front of the fraction.

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

Step 6

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

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