Calculus Examples

$5 x y + y - x - 3 = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (5 x y + y - x - 3) = \frac{d}{d x} (0)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $5 x y + y - x - 3$ with respect to $x$ is $\frac{d}{d x} [5 x y] + \frac{d}{d x} [y] + \frac{d}{d x} [- x] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [5 x y] + \frac{d}{d x} [y] + \frac{d}{d x} [- x] + \frac{d}{d x} [- 3]$

Step 2.2

Evaluate $\frac{d}{d x} [5 x y]$ .

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Step 2.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x y$ with respect to $x$ is $5 \frac{d}{d x} [x y]$ .

$5 \frac{d}{d x} [x y] + \frac{d}{d x} [y] + \frac{d}{d x} [- x] + \frac{d}{d x} [- 3]$

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 $g (x) = y$ .

$5 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x]) + \frac{d}{d x} [y] + \frac{d}{d x} [- x] + \frac{d}{d x} [- 3]$

Step 2.2.3

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

$5 (x y' + y \frac{d}{d x} [x]) + \frac{d}{d x} [y] + \frac{d}{d x} [- x] + \frac{d}{d x} [- 3]$

Step 2.2.4

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

$5 (x y' + y \cdot 1) + \frac{d}{d x} [y] + \frac{d}{d x} [- x] + \frac{d}{d x} [- 3]$

Step 2.2.5

Multiply $y$ by $1$ .

$5 (x y' + y) + \frac{d}{d x} [y] + \frac{d}{d x} [- x] + \frac{d}{d x} [- 3]$

Step 2.3

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

$5 (x y' + y) + y' + \frac{d}{d x} [- x] + \frac{d}{d x} [- 3]$

Step 2.4

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

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Step 2.4.1

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

$5 (x y' + y) + y' - \frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

Step 2.4.2

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

$5 (x y' + y) + y' - 1 \cdot 1 + \frac{d}{d x} [- 3]$

Step 2.4.3

Multiply $- 1$ by $1$ .

$5 (x y' + y) + y' - 1 + \frac{d}{d x} [- 3]$

Step 2.5

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

$5 (x y' + y) + y' - 1 + 0$

Step 2.6

Simplify.

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Step 2.6.1

Apply the distributive property.

$5 (x y') + 5 y + y' - 1 + 0$

Step 2.6.2

Add $5 x y' + 5 y + y' - 1$ and $0$ .

$5 x y' + 5 y + y' - 1$

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.

$5 x y' + 5 y + y' - 1 = 0$

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 $5 y$ from both sides of the equation.

$5 x y' + y' - 1 = - 5 y$

Step 5.1.2

Add $1$ to both sides of the equation.

$5 x y' + y' = - 5 y + 1$

Step 5.2

Factor $y'$ out of $5 x y' + y'$ .

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Step 5.2.1

Factor $y'$ out of $5 x y'$ .

$y' (5 x) + y' = - 5 y + 1$

Step 5.2.2

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

$y' (5 x) + y' = - 5 y + 1$

Step 5.2.3

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

$y' (5 x) + y' \cdot 1 = - 5 y + 1$

Step 5.2.4

Factor $y'$ out of $y' (5 x) + y' \cdot 1$ .

$y' (5 x + 1) = - 5 y + 1$

Step 5.3

Divide each term in $y' (5 x + 1) = - 5 y + 1$ by $5 x + 1$ and simplify.

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Step 5.3.1

Divide each term in $y' (5 x + 1) = - 5 y + 1$ by $5 x + 1$ .

$\frac{y' (5 x + 1)}{5 x + 1} = \frac{- 5 y}{5 x + 1} + \frac{1}{5 x + 1}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $5 x + 1$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{y' (5 x + 1)}{5 x + 1} = \frac{- 5 y}{5 x + 1} + \frac{1}{5 x + 1}$

Step 5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 5 y}{5 x + 1} + \frac{1}{5 x + 1}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

$y' = \frac{- 5 y + 1}{5 x + 1}$

Step 5.3.3.2

Factor $- 1$ out of $- 5 y$ .

$y' = \frac{- (5 y) + 1}{5 x + 1}$

Step 5.3.3.3

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

$y' = \frac{- (5 y) - 1 (- 1)}{5 x + 1}$

Step 5.3.3.4

Factor $- 1$ out of $- (5 y) - 1 (- 1)$ .

$y' = \frac{- (5 y - 1)}{5 x + 1}$

Step 5.3.3.5

Simplify the expression.

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Step 5.3.3.5.1

Rewrite $- (5 y - 1)$ as $- 1 (5 y - 1)$ .

$y' = \frac{- 1 (5 y - 1)}{5 x + 1}$

Step 5.3.3.5.2

Move the negative in front of the fraction.

$y' = - \frac{5 y - 1}{5 x + 1}$

Step 6

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

$\frac{d y}{d x} = - \frac{5 y - 1}{5 x + 1}$