Calculus Examples

$1 - x y = x - y$

Step 1

Differentiate both sides of the equation.

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

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

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

Step 2.1.2

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

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

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

$0 - \frac{d}{d x} [x y]$

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 $g (x) = y$ .

$0 - (x \frac{d}{d x} [y] + y \frac{d}{d x} [x])$

Step 2.2.3

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

$0 - (x y' + y \frac{d}{d x} [x])$

Step 2.2.4

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

$0 - (x y' + y \cdot 1)$

Step 2.2.5

Multiply $y$ by $1$ .

$0 - (x y' + y)$

Step 2.3

Simplify.

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Step 2.3.1

Apply the distributive property.

$0 - (x y') - y$

Step 2.3.2

Subtract $- (- x y' - y)$ from $0$ .

$- x y' - y$

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

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

Step 3.1.2

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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

$1 - y'$

Step 4

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

$- x y' - y = 1 - y'$

Step 5

Solve for $y'$ .

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Step 5.1

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

$- x y' - y + y' = 1$

Step 5.2

Add $y$ to both sides of the equation.

$- x y' + y' = 1 + y$

Step 5.3

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

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Step 5.3.1

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

$y' (- x) + y' = 1 + y$

Step 5.3.2

Raise $y'$ to the power of $1$ .

$y' (- x) + y' = 1 + y$

Step 5.3.3

Factor $y'$ out of ${y'}^{1}$ .

$y' (- x) + y' \cdot 1 = 1 + y$

Step 5.3.4

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

$y' (- x + 1) = 1 + y$

Step 5.4

Divide each term in $y' (- x + 1) = 1 + y$ by $- x + 1$ and simplify.

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Step 5.4.1

Divide each term in $y' (- x + 1) = 1 + y$ by $- x + 1$ .

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

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

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Divide $y'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Combine the numerators over the common denominator.

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

Step 5.4.3.2

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

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

Step 5.4.3.3

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

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

Step 5.4.3.4

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

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

Step 5.4.3.5

Rewrite negatives.

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Step 5.4.3.5.1

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

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

Step 5.4.3.5.2

Move the negative in front of the fraction.

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

Step 6

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

$\frac{d y}{d x} = - \frac{1 + y}{x - 1}$