Calculus Examples

$x^{2} - 8 x y + y^{2} = 8$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{2} - 8 x y + y^{2}) = \frac{d}{d x} (8)$

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 $x^{2} - 8 x y + y^{2}$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 8 x y] + \frac{d}{d x} [y^{2}]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 8 x y] + \frac{d}{d x} [y^{2}]$

Step 2.1.2

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

$2 x + \frac{d}{d x} [- 8 x y] + \frac{d}{d x} [y^{2}]$

Step 2.2

Evaluate $\frac{d}{d x} [- 8 x y]$ .

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Step 2.2.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 x y$ with respect to $x$ is $- 8 \frac{d}{d x} [x y]$ .

$2 x - 8 \frac{d}{d x} [x y] + \frac{d}{d x} [y^{2}]$

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 $g (x) = y$ .

$2 x - 8 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x]) + \frac{d}{d x} [y^{2}]$

Step 2.2.3

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

$2 x - 8 (x y' + y \frac{d}{d x} [x]) + \frac{d}{d x} [y^{2}]$

Step 2.2.4

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

$2 x - 8 (x y' + y \cdot 1) + \frac{d}{d x} [y^{2}]$

Step 2.2.5

Multiply $y$ by $1$ .

$2 x - 8 (x y' + y) + \frac{d}{d x} [y^{2}]$

Step 2.3

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

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Step 2.3.1

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 2.3.1.1

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

$2 x - 8 (x y' + y) + \frac{d}{d u} [u^{2}] \frac{d}{d x} [y]$

Step 2.3.1.2

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

$2 x - 8 (x y' + y) + 2 u \frac{d}{d x} [y]$

Step 2.3.1.3

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

$2 x - 8 (x y' + y) + 2 y \frac{d}{d x} [y]$

Step 2.3.2

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

$2 x - 8 (x y' + y) + 2 y y'$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$2 x - 8 (x y') - 8 y + 2 y y'$

Step 2.4.2

Remove unnecessary parentheses.

$2 x - 8 x y' - 8 y + 2 y y'$

Step 2.4.3

Reorder terms.

$- 8 x y' + 2 y y' + 2 x - 8 y$

Step 3

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$0$

Step 4

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

$- 8 x y' + 2 y y' + 2 x - 8 y = 0$

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

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

$- 8 x y' + 2 y y' - 8 y = - 2 x$

Step 5.1.2

Add $8 y$ to both sides of the equation.

$- 8 x y' + 2 y y' = - 2 x + 8 y$

Step 5.2

Factor $2 y'$ out of $- 8 x y' + 2 y y'$ .

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Step 5.2.1

Factor $2 y'$ out of $- 8 x y'$ .

$2 y' (- 4 x) + 2 y y' = - 2 x + 8 y$

Step 5.2.2

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

$2 y' (- 4 x) + 2 y' y = - 2 x + 8 y$

Step 5.2.3

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

$2 y' (- 4 x + y) = - 2 x + 8 y$

Step 5.3

Divide each term in $2 y' (- 4 x + y) = - 2 x + 8 y$ by $2 (- 4 x + y)$ and simplify.

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Step 5.3.1

Divide each term in $2 y' (- 4 x + y) = - 2 x + 8 y$ by $2 (- 4 x + y)$ .

$\frac{2 y' (- 4 x + y)}{2 (- 4 x + y)} = \frac{- 2 x}{2 (- 4 x + y)} + \frac{8 y}{2 (- 4 x + y)}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $2$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{2 y' (- 4 x + y)}{2 (- 4 x + y)} = \frac{- 2 x}{2 (- 4 x + y)} + \frac{8 y}{2 (- 4 x + y)}$

Step 5.3.2.1.2

Rewrite the expression.

$\frac{y' (- 4 x + y)}{- 4 x + y} = \frac{- 2 x}{2 (- 4 x + y)} + \frac{8 y}{2 (- 4 x + y)}$

Step 5.3.2.2

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

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Step 5.3.2.2.1

Cancel the common factor.

$\frac{y' (- 4 x + y)}{- 4 x + y} = \frac{- 2 x}{2 (- 4 x + y)} + \frac{8 y}{2 (- 4 x + y)}$

Step 5.3.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 2 x}{2 (- 4 x + y)} + \frac{8 y}{2 (- 4 x + y)}$

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 $- 2$ and $2$ .

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Step 5.3.3.1.1.1

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

$y' = \frac{2 (- x)}{2 (- 4 x + y)} + \frac{8 y}{2 (- 4 x + y)}$

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.

$y' = \frac{2 (- x)}{2 (- 4 x + y)} + \frac{8 y}{2 (- 4 x + y)}$

Step 5.3.3.1.1.2.2

Rewrite the expression.

$y' = \frac{- x}{- 4 x + y} + \frac{8 y}{2 (- 4 x + y)}$

Step 5.3.3.1.2

Move the negative in front of the fraction.

$y' = - \frac{x}{- 4 x + y} + \frac{8 y}{2 (- 4 x + y)}$

Step 5.3.3.1.3

Cancel the common factor of $8$ and $2$ .

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Step 5.3.3.1.3.1

Factor $2$ out of $8 y$ .

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

Step 5.3.3.1.3.2

Cancel the common factors.

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Step 5.3.3.1.3.2.1

Cancel the common factor.

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

Step 5.3.3.1.3.2.2

Rewrite the expression.

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

Step 5.3.3.2

Simplify terms.

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Step 5.3.3.2.1

Combine the numerators over the common denominator.

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

Step 5.3.3.2.2

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

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

Step 5.3.3.2.3

Factor $- 1$ out of $4 y$ .

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

Step 5.3.3.2.4

Factor $- 1$ out of $- (x) - (- 4 y)$ .

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

Step 5.3.3.2.5

Rewrite $- (x - 4 y)$ as $- 1 (x - 4 y)$ .

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

Step 5.3.3.2.6

Factor $- 1$ out of $- 4 x$ .

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

Step 5.3.3.2.7

Factor $- 1$ out of $y$ .

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

Step 5.3.3.2.8

Factor $- 1$ out of $- (4 x) - 1 (- y)$ .

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

Step 5.3.3.2.9

Rewrite $- (4 x - y)$ as $- 1 (4 x - y)$ .

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

Step 5.3.3.2.10

Cancel the common factor.

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

Step 5.3.3.2.11

Rewrite the expression.

$y' = \frac{x - 4 y}{4 x - y}$

Step 6

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

$\frac{d y}{d x} = \frac{x - 4 y}{4 x - y}$