Calculus Examples

$3 y^{3} + x y - 6 x + 3 = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (3 y^{3} + x y - 6 x + 3) = \frac{d}{d x} (0)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $3 y^{3} + x y - 6 x + 3$ with respect to $x$ is $\frac{d}{d x} [3 y^{3}] + \frac{d}{d x} [x y] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [3]$ .

$\frac{d}{d x} [3 y^{3}] + \frac{d}{d x} [x y] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [3]$

Step 2.2

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

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Step 2.2.1

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

$3 \frac{d}{d x} [y^{3}] + \frac{d}{d x} [x y] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [3]$

Step 2.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = y$ .

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Step 2.2.2.1

To apply the Chain Rule, set $u$ as $y$ .

$3 (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y]) + \frac{d}{d x} [x y] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [3]$

Step 2.2.2.2

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

$3 (3 u^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [x y] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [3]$

Step 2.2.2.3

Replace all occurrences of $u$ with $y$ .

$3 (3 y^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [x y] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [3]$

Step 2.2.3

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

$3 (3 y^{2} y') + \frac{d}{d x} [x y] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [3]$

Step 2.2.4

Multiply $3$ by $3$ .

$9 y^{2} y' + \frac{d}{d x} [x y] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [3]$

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 $g (x) = y$ .

$9 y^{2} y' + x \frac{d}{d x} [y] + y \frac{d}{d x} [x] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [3]$

Step 2.3.2

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

$9 y^{2} y' + x y' + y \frac{d}{d x} [x] + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [3]$

Step 2.3.3

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

$9 y^{2} y' + x y' + y \cdot 1 + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [3]$

Step 2.3.4

Multiply $y$ by $1$ .

$9 y^{2} y' + x y' + y + \frac{d}{d x} [- 6 x] + \frac{d}{d x} [3]$

Step 2.4

Evaluate $\frac{d}{d x} [- 6 x]$ .

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Step 2.4.1

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 x$ with respect to $x$ is $- 6 \frac{d}{d x} [x]$ .

$9 y^{2} y' + x y' + y - 6 \frac{d}{d x} [x] + \frac{d}{d x} [3]$

Step 2.4.2

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

$9 y^{2} y' + x y' + y - 6 \cdot 1 + \frac{d}{d x} [3]$

Step 2.4.3

Multiply $- 6$ by $1$ .

$9 y^{2} y' + x y' + y - 6 + \frac{d}{d x} [3]$

Step 2.5

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

$9 y^{2} y' + x y' + y - 6 + 0$

Step 2.6

Add $9 y^{2} y' + x y' + y - 6$ and $0$ .

$9 y^{2} y' + x y' + y - 6$

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.

$9 y^{2} y' + x y' + y - 6 = 0$

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 $y$ from both sides of the equation.

$9 y^{2} y' + x y' - 6 = - y$

Step 5.1.2

Add $6$ to both sides of the equation.

$9 y^{2} y' + x y' = - y + 6$

Step 5.2

Factor $y'$ out of $9 y^{2} y' + x y'$ .

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Step 5.2.1

Factor $y'$ out of $9 y^{2} y'$ .

$y' (9 y^{2}) + x (y') = - y + 6$

Step 5.2.2

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

$y' (9 y^{2}) + y' (x) = - y + 6$

Step 5.2.3

Factor $y'$ out of $y' (9 y^{2}) + y' (x)$ .

$y' (9 y^{2} + x) = - y + 6$

Step 5.3

Divide each term in $y' (9 y^{2} + x) = - y + 6$ by $9 y^{2} + x$ and simplify.

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Step 5.3.1

Divide each term in $y' (9 y^{2} + x) = - y + 6$ by $9 y^{2} + x$ .

$\frac{y' (9 y^{2} + x)}{9 y^{2} + x} = \frac{- y}{9 y^{2} + x} + \frac{6}{9 y^{2} + x}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $9 y^{2} + x$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{y' (9 y^{2} + x)}{9 y^{2} + x} = \frac{- y}{9 y^{2} + x} + \frac{6}{9 y^{2} + x}$

Step 5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- y}{9 y^{2} + x} + \frac{6}{9 y^{2} + x}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

$y' = \frac{- y + 6}{9 y^{2} + x}$

Step 5.3.3.2

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

$y' = \frac{- (y) + 6}{9 y^{2} + x}$

Step 5.3.3.3

Rewrite $6$ as $- 1 (- 6)$ .

$y' = \frac{- (y) - 1 (- 6)}{9 y^{2} + x}$

Step 5.3.3.4

Factor $- 1$ out of $- (y) - 1 (- 6)$ .

$y' = \frac{- (y - 6)}{9 y^{2} + x}$

Step 5.3.3.5

Simplify the expression.

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Step 5.3.3.5.1

Rewrite $- (y - 6)$ as $- 1 (y - 6)$ .

$y' = \frac{- 1 (y - 6)}{9 y^{2} + x}$

Step 5.3.3.5.2

Move the negative in front of the fraction.

$y' = - \frac{y - 6}{9 y^{2} + x}$

Step 6

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

$\frac{d y}{d x} = - \frac{y - 6}{9 y^{2} + x}$