Calculus Examples

$x^{2} y + y^{4} = 4 + 2 x$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (x^{2} y + y^{4}) = \frac{d}{d y} (4 + 2 x)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $x^{2} y + y^{4}$ with respect to $y$ is $\frac{d}{d y} [x^{2} y] + \frac{d}{d y} [y^{4}]$ .

$\frac{d}{d y} [x^{2} y] + \frac{d}{d y} [y^{4}]$

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 $g (y) = y$ .

$x^{2} \frac{d}{d y} [y] + y \frac{d}{d y} [x^{2}] + \frac{d}{d y} [y^{4}]$

Step 2.2.2

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

$x^{2} \cdot 1 + y \frac{d}{d y} [x^{2}] + \frac{d}{d y} [y^{4}]$

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 $x$ .

$x^{2} \cdot 1 + y (\frac{d}{d u} [u^{2}] \frac{d}{d y} [x]) + \frac{d}{d y} [y^{4}]$

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 $n = 2$ .

$x^{2} \cdot 1 + y (2 u \frac{d}{d y} [x]) + \frac{d}{d y} [y^{4}]$

Step 2.2.3.3

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

$x^{2} \cdot 1 + y (2 x \frac{d}{d y} [x]) + \frac{d}{d y} [y^{4}]$

Step 2.2.4

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

$x^{2} \cdot 1 + y (2 x x') + \frac{d}{d y} [y^{4}]$

Step 2.2.5

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

$x^{2} + y (2 x x') + \frac{d}{d y} [y^{4}]$

Step 2.2.6

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

$x^{2} + 2 y x x' + \frac{d}{d y} [y^{4}]$

Step 2.3

Differentiate using the Power Rule.

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Step 2.3.1

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 4$ .

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

Step 2.3.2

Reorder terms.

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate.

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Step 3.1.1

By the Sum Rule, the derivative of $4 + 2 x$ with respect to $y$ is $\frac{d}{d y} [4] + \frac{d}{d y} [2 x]$ .

$\frac{d}{d y} [4] + \frac{d}{d y} [2 x]$

Step 3.1.2

Since $4$ is constant with respect to $y$ , the derivative of $4$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [2 x]$

Step 3.2

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

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Step 3.2.1

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

$0 + 2 \frac{d}{d y} [x]$

Step 3.2.2

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

$0 + 2 x'$

Step 3.3

Add $0$ and $2 x'$ .

$2 x'$

Step 4

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

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

Step 5

Solve for $x'$ .

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Step 5.1

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

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

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 $4 y^{3}$ from both sides of the equation.

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

Step 5.2.2

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 5.3.3

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

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

Step 5.4

Divide each term in $2 x' (x y - 1) = - 4 y^{3} - x^{2}$ by $2 (x y - 1)$ and simplify.

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Step 5.4.1

Divide each term in $2 x' (x y - 1) = - 4 y^{3} - x^{2}$ by $2 (x y - 1)$ .

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $2$ .

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Rewrite the expression.

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

Step 5.4.2.2

Cancel the common factor of $x y - 1$ .

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Step 5.4.2.2.1

Cancel the common factor.

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

Step 5.4.2.2.2

Divide $x'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Simplify each term.

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Step 5.4.3.1.1

Cancel the common factor of $- 4$ and $2$ .

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Step 5.4.3.1.1.1

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

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

Step 5.4.3.1.1.2

Cancel the common factors.

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Step 5.4.3.1.1.2.1

Cancel the common factor.

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

Step 5.4.3.1.1.2.2

Rewrite the expression.

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

Step 5.4.3.1.2

Move the negative in front of the fraction.

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

Step 5.4.3.1.3

Move the negative in front of the fraction.

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

Step 5.4.3.2

To write $- \frac{2 y^{3}}{x y - 1}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 5.4.3.3

Write each expression with a common denominator of $(x y - 1) \cdot 2$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.4.3.3.1

Multiply $\frac{2 y^{3}}{x y - 1}$ by $\frac{2}{2}$ .

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

Step 5.4.3.3.2

Reorder the factors of $(x y - 1) \cdot 2$ .

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

Step 5.4.3.4

Combine the numerators over the common denominator.

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

Step 5.4.3.5

Multiply $2$ by $- 2$ .

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

Step 5.4.3.6

Factor $- 1$ out of $- 4 y^{3}$ .

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

Step 5.4.3.7

Factor $- 1$ out of $- x^{2}$ .

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

Step 5.4.3.8

Factor $- 1$ out of $- (4 y^{3}) - (x^{2})$ .

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

Step 5.4.3.9

Simplify the expression.

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Step 5.4.3.9.1

Rewrite $- (4 y^{3} + x^{2})$ as $- 1 (4 y^{3} + x^{2})$ .

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

Step 5.4.3.9.2

Move the negative in front of the fraction.

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

Step 6

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

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

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