Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

Differentiate.

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.2

Evaluate $\frac{d}{d x} [- 4 x y]$ .

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Step 2.2.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x y$ with respect to $x$ is $- 4 \frac{d}{d x} [x y]$ .

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

Step 2.2.2

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 $g (x) = y$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Multiply $y$ by $1$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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 2.3.2.1

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

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

Step 2.3.2.2

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

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

Step 2.3.2.3

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

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

Step 2.3.3

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

$2 x - 4 (x y' + y) + 2 (2 y y')$

Step 2.3.4

Multiply $2$ by $2$ .

$2 x - 4 (x y' + y) + 4 y y'$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$2 x - 4 (x y') - 4 y + 4 y y'$

Step 2.4.2

Remove unnecessary parentheses.

$2 x - 4 x y' - 4 y + 4 y y'$

Step 2.4.3

Reorder terms.

$- 4 x y' + 4 y y' + 2 x - 4 y$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$- 4 \frac{d}{d x} [x]$

Step 3.2

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

$- 4 \cdot 1$

Step 3.3

Multiply $- 4$ by $1$ .

$- 4$

Step 4

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

$- 4 x y' + 4 y y' + 2 x - 4 y = - 4$

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

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

$- 4 x y' + 4 y y' - 4 y = - 4 - 2 x$

Step 5.1.2

Add $4 y$ to both sides of the equation.

$- 4 x y' + 4 y y' = - 4 - 2 x + 4 y$

Step 5.2

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

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Step 5.2.1

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

$4 y' (- x) + 4 y y' = - 4 - 2 x + 4 y$

Step 5.2.2

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

$4 y' (- x) + 4 y' y = - 4 - 2 x + 4 y$

Step 5.2.3

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

$4 y' (- x + y) = - 4 - 2 x + 4 y$

Step 5.3

Divide each term in $4 y' (- x + y) = - 4 - 2 x + 4 y$ by $4 (- x + y)$ and simplify.

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Step 5.3.1

Divide each term in $4 y' (- x + y) = - 4 - 2 x + 4 y$ by $4 (- x + y)$ .

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $4$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Rewrite the expression.

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

Step 5.3.2.2

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

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Step 5.3.2.2.1

Cancel the common factor.

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

Step 5.3.2.2.2

Divide $y'$ by $1$ .

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

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 $- 4$ and $4$ .

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Step 5.3.3.1.1.1

Factor $4$ out of $- 4$ .

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

Step 5.3.3.1.1.2

Cancel the common factors.

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Step 5.3.3.1.1.2.1

Cancel the common factor.

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

Step 5.3.3.1.1.2.2

Rewrite the expression.

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

Step 5.3.3.1.2

Move the negative in front of the fraction.

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

Step 5.3.3.1.3

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

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Step 5.3.3.1.3.1

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

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

Step 5.3.3.1.3.2

Cancel the common factors.

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Step 5.3.3.1.3.2.1

Factor $2$ out of $4 (- x + y)$ .

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

Step 5.3.3.1.3.2.2

Cancel the common factor.

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

Step 5.3.3.1.3.2.3

Rewrite the expression.

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

Step 5.3.3.1.4

Move the negative in front of the fraction.

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

Step 5.3.3.1.5

Cancel the common factor of $4$ .

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Step 5.3.3.1.5.1

Cancel the common factor.

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

Step 5.3.3.1.5.2

Rewrite the expression.

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

Step 5.3.3.2

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

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

Step 5.3.3.3

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

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Step 5.3.3.3.1

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

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

Step 5.3.3.3.2

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

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

Step 5.3.3.4

Combine the numerators over the common denominator.

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

Step 5.3.3.5

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

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

Step 5.3.3.6

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

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Step 5.3.3.6.1

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

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

Step 5.3.3.6.2

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

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

Step 5.3.3.7

Combine the numerators over the common denominator.

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

Step 5.3.3.8

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

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

Step 5.3.3.9

Simplify terms.

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Step 5.3.3.9.1

Rewrite $- 2$ as $- 1 (2)$ .

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

Step 5.3.3.9.2

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

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

Step 5.3.3.9.3

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

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

Step 5.3.3.9.4

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

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

Step 5.3.3.9.5

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

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

Step 5.3.3.9.6

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

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

Step 5.3.3.9.7

Factor $- 1$ out of $y$ .

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

Step 5.3.3.9.8

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

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

Step 5.3.3.9.9

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

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

Step 5.3.3.9.10

Cancel the common factor.

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

Step 5.3.3.9.11

Rewrite the expression.

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

Step 6

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

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