Calculus Examples

$- x y + 3 = y$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $- x y + 3$ with respect to $x$ is $\frac{d}{d x} [- x y] + \frac{d}{d x} [3]$ .

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

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

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

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 $g (x) = y$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Multiply $y$ by $1$ .

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

Step 2.3

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

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

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

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

Step 2.4.2

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

$- x y' - y$

Step 3

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

$y'$

Step 4

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

$- x y' - y = y'$

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $y'$ from both sides of the equation.

$- x y' - y - y' = 0$

Step 5.2

Add $y$ to both sides of the equation.

$- x y' - y' = y$

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 5.3.3

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

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

Step 5.4

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

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Step 5.4.1

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

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

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.

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

Step 5.4.2.1.2

Divide $y'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

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

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

Step 5.4.3.2

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

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

Step 5.4.3.3

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

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

Step 5.4.3.4

Rewrite negatives.

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Step 5.4.3.4.1

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

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

Step 5.4.3.4.2

Move the negative in front of the fraction.

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

Step 6

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

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