Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

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

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

Step 2.1.2

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

$0 + \frac{d}{d x} [- x^{3}] + \frac{d}{d x} [x y] + \frac{d}{d x} [3 y]$

Step 2.2

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

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Step 2.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{3}$ with respect to $x$ is $- \frac{d}{d x} [x^{3}]$ .

$0 - \frac{d}{d x} [x^{3}] + \frac{d}{d x} [x y] + \frac{d}{d x} [3 y]$

Step 2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$0 - (3 x^{2}) + \frac{d}{d x} [x y] + \frac{d}{d x} [3 y]$

Step 2.2.3

Multiply $3$ by $- 1$ .

$0 - 3 x^{2} + \frac{d}{d x} [x y] + \frac{d}{d x} [3 y]$

Step 2.3

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

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Step 2.3.1

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 $g (x) = y$ .

$0 - 3 x^{2} + x \frac{d}{d x} [y] + y \frac{d}{d x} [x] + \frac{d}{d x} [3 y]$

Step 2.3.2

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

$0 - 3 x^{2} + x y' + y \frac{d}{d x} [x] + \frac{d}{d x} [3 y]$

Step 2.3.3

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

$0 - 3 x^{2} + x y' + y \cdot 1 + \frac{d}{d x} [3 y]$

Step 2.3.4

Multiply $y$ by $1$ .

$0 - 3 x^{2} + x y' + y + \frac{d}{d x} [3 y]$

Step 2.4

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

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Step 2.4.1

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

$0 - 3 x^{2} + x y' + y + 3 \frac{d}{d x} [y]$

Step 2.4.2

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

$0 - 3 x^{2} + x y' + y + 3 y'$

Step 2.5

Subtract $3 x^{2}$ from $0$ .

$- 3 x^{2} + x y' + y + 3 y'$

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.

$- 3 x^{2} + x y' + y + 3 y' = 0$

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

Add $3 x^{2}$ to both sides of the equation.

$x y' + y + 3 y' = 3 x^{2}$

Step 5.1.2

Subtract $y$ from both sides of the equation.

$x y' + 3 y' = 3 x^{2} - y$

Step 5.2

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

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Step 5.2.1

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

$y' x + 3 y' = 3 x^{2} - y$

Step 5.2.2

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

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

Step 5.2.3

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

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

Step 5.3

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

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Step 5.3.1

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

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

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.

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

Step 5.3.2.1.2

Divide $y'$ by $1$ .

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

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.

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

Step 5.3.3.2

Combine the numerators over the common denominator.

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

Step 6

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

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