Calculus Examples

$x^{2} - x y - 3 y - 1 = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{2} - x y - 3 y - 1) = \frac{d}{d x} (0)$

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 $x^{2} - x y - 3 y - 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- x y] + \frac{d}{d x} [- 3 y] + \frac{d}{d x} [- 1]$ .

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

Step 2.1.2

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

$2 x + \frac{d}{d x} [- x y] + \frac{d}{d x} [- 3 y] + \frac{d}{d x} [- 1]$

Step 2.2

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

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Step 2.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x y$ with respect to $x$ is $- \frac{d}{d x} [x y]$ .

$2 x - \frac{d}{d x} [x y] + \frac{d}{d x} [- 3 y] + \frac{d}{d x} [- 1]$

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 $g (x) = y$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Multiply $y$ by $1$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$2 x - (x y' + y) - 3 y' + 0$

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

$2 x - (x y') - y - 3 y' + 0$

Step 2.5.2

Add $2 x - x y' - y - 3 y'$ and $0$ .

$2 x - x y' - y - 3 y'$

Step 2.5.3

Reorder terms.

$- x y' + 2 x - y - 3 y'$

Step 3

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

$0$

Step 4

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

$- x y' + 2 x - y - 3 y' = 0$

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.

$- x y' - y - 3 y' = - 2 x$

Step 5.1.2

Add $y$ to both sides of the equation.

$- x y' - 3 y' = - 2 x + y$

Step 5.2

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

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Step 5.2.1

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

$y' (- x) - 3 y' = - 2 x + y$

Step 5.2.2

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

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

Step 5.2.3

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

$y' (- x - 3) = - 2 x + y$

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

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

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Divide $y'$ by $1$ .

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Move the negative in front of the fraction.

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

Step 5.3.3.2

Combine the numerators over the common denominator.

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

Step 5.3.3.3

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

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

Step 5.3.3.4

Factor $- 1$ out of $y$ .

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

Step 5.3.3.5

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

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

Step 5.3.3.6

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

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

Step 5.3.3.7

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

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

Step 5.3.3.8

Rewrite $- 3$ as $- 1 (3)$ .

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

Step 5.3.3.9

Factor $- 1$ out of $- (x) - 1 (3)$ .

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

Step 5.3.3.10

Rewrite $- (x + 3)$ as $- 1 (x + 3)$ .

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

Step 5.3.3.11

Cancel the common factor.

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

Step 5.3.3.12

Rewrite the expression.

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

Step 6

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

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