Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $3$ by $- 3$ .

$- 9 x^{2} + \frac{d}{d x} [x y] + \frac{d}{d x} [- 3]$

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 $g (x) = y$ .

$- 9 x^{2} + x \frac{d}{d x} [y] + y \frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

Step 2.3.2

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

$- 9 x^{2} + x y' + y \frac{d}{d x} [x] + \frac{d}{d x} [- 3]$

Step 2.3.3

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

$- 9 x^{2} + x y' + y \cdot 1 + \frac{d}{d x} [- 3]$

Step 2.3.4

Multiply $y$ by $1$ .

$- 9 x^{2} + x y' + y + \frac{d}{d x} [- 3]$

Step 2.4

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

$- 9 x^{2} + x y' + y + 0$

Step 2.5

Add $- 9 x^{2} + x y' + y$ and $0$ .

$- 9 x^{2} + x y' + y$

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.

$- 9 x^{2} + x y' + y = 0$

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 $9 x^{2}$ to both sides of the equation.

$x y' + y = 9 x^{2}$

Step 5.1.2

Subtract $y$ from both sides of the equation.

$x y' = 9 x^{2} - y$

Step 5.2

Divide each term in $x y' = 9 x^{2} - y$ by $x$ and simplify.

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Step 5.2.1

Divide each term in $x y' = 9 x^{2} - y$ by $x$ .

$\frac{x y'}{x} = \frac{9 x^{2}}{x} + \frac{- y}{x}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $x$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{x y'}{x} = \frac{9 x^{2}}{x} + \frac{- y}{x}$

Step 5.2.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{9 x^{2}}{x} + \frac{- y}{x}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Simplify each term.

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Step 5.2.3.1.1

Cancel the common factor of $x^{2}$ and $x$ .

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Step 5.2.3.1.1.1

Factor $x$ out of $9 x^{2}$ .

$y' = \frac{x (9 x)}{x} + \frac{- y}{x}$

Step 5.2.3.1.1.2

Cancel the common factors.

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Step 5.2.3.1.1.2.1

Raise $x$ to the power of $1$ .

$y' = \frac{x (9 x)}{x^{1}} + \frac{- y}{x}$

Step 5.2.3.1.1.2.2

Factor $x$ out of $x^{1}$ .

$y' = \frac{x (9 x)}{x \cdot 1} + \frac{- y}{x}$

Step 5.2.3.1.1.2.3

Cancel the common factor.

$y' = \frac{x (9 x)}{x \cdot 1} + \frac{- y}{x}$

Step 5.2.3.1.1.2.4

Rewrite the expression.

$y' = \frac{9 x}{1} + \frac{- y}{x}$

Step 5.2.3.1.1.2.5

Divide $9 x$ by $1$ .

$y' = 9 x + \frac{- y}{x}$

Step 5.2.3.1.2

Move the negative in front of the fraction.

$y' = 9 x - \frac{y}{x}$

Step 6

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

$\frac{d y}{d x} = 9 x - \frac{y}{x}$