Calculus Examples

$y + 3 x y - 5 = 0$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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 $g (x) = y$ .

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

Multiply $y$ by $1$ .

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

Step 2.4

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

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

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

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

Step 2.5.2

Add $y' + 3 x y' + 3 y$ and $0$ .

$y' + 3 x y' + 3 y$

Step 2.5.3

Reorder terms.

$3 x y' + 3 y + y'$

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.

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

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $3 y$ from both sides of the equation.

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.2.3

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

$y' (3 x) + y' \cdot 1 = - 3 y$

Step 5.2.4

Factor $y'$ out of $y' (3 x) + y' \cdot 1$ .

$y' (3 x + 1) = - 3 y$

Step 5.3

Divide each term in $y' (3 x + 1) = - 3 y$ by $3 x + 1$ and simplify.

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Step 5.3.1

Divide each term in $y' (3 x + 1) = - 3 y$ by $3 x + 1$ .

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $3 x + 1$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Divide $y'$ by $1$ .

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

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.

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

Step 6

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

$\frac{d y}{d x} = - \frac{3 y}{3 x + 1}$