Calculus Examples

$y^{3} x + x + 2 y = 8$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y^{3} x + x + 2 y) = \frac{d}{d x} (8)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $y^{3} x + x + 2 y$ with respect to $x$ is $\frac{d}{d x} [y^{3} x] + \frac{d}{d x} [x] + \frac{d}{d x} [2 y]$ .

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

Step 2.2

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

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Step 2.2.1

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 $f (x) = y^{3}$ and $g (x) = x$ .

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

Step 2.2.2

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

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

Step 2.2.3

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 2.2.3.1

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

$y^{3} \cdot 1 + x (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y]) + \frac{d}{d x} [x] + \frac{d}{d x} [2 y]$

Step 2.2.3.2

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

$y^{3} \cdot 1 + x (3 u^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [x] + \frac{d}{d x} [2 y]$

Step 2.2.3.3

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

$y^{3} \cdot 1 + x (3 y^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [x] + \frac{d}{d x} [2 y]$

Step 2.2.4

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

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

Step 2.2.5

Multiply $y^{3}$ by $1$ .

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

Step 2.2.6

Move $3$ to the left of $x$ .

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

Step 2.3

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

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

Step 2.4

Evaluate $\frac{d}{d x} [2 y]$ .

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Step 2.4.1

Since $2$ is constant with respect to $x$ , the derivative of $2 y$ with respect to $x$ is $2 \frac{d}{d x} [y]$ .

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

Step 2.4.2

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

$y^{3} + 3 x y^{2} y' + 1 + 2 y'$

Step 2.5

Reorder terms.

$y^{3} + 3 y^{2} x y' + 2 y' + 1$

Step 3

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$0$

Step 4

Reform the equation by setting the left side equal to the right side.

$y^{3} + 3 y^{2} x y' + 2 y' + 1 = 0$

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 $y^{3}$ from both sides of the equation.

$3 y^{2} x y' + 2 y' + 1 = - y^{3}$

Step 5.1.2

Subtract $1$ from both sides of the equation.

$3 y^{2} x y' + 2 y' = - y^{3} - 1$

Step 5.2

Factor $y'$ out of $3 y^{2} x y' + 2 y'$ .

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Step 5.2.1

Factor $y'$ out of $3 y^{2} x y'$ .

$y' (3 y^{2} x) + 2 y' = - y^{3} - 1$

Step 5.2.2

Factor $y'$ out of $2 y'$ .

$y' (3 y^{2} x) + y' \cdot 2 = - y^{3} - 1$

Step 5.2.3

Factor $y'$ out of $y' (3 y^{2} x) + y' \cdot 2$ .

$y' (3 y^{2} x + 2) = - y^{3} - 1$

Step 5.3

Divide each term in $y' (3 y^{2} x + 2) = - y^{3} - 1$ by $3 y^{2} x + 2$ and simplify.

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Step 5.3.1

Divide each term in $y' (3 y^{2} x + 2) = - y^{3} - 1$ by $3 y^{2} x + 2$ .

$\frac{y' (3 y^{2} x + 2)}{3 y^{2} x + 2} = \frac{- y^{3}}{3 y^{2} x + 2} + \frac{- 1}{3 y^{2} x + 2}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $3 y^{2} x + 2$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{y' (3 y^{2} x + 2)}{3 y^{2} x + 2} = \frac{- y^{3}}{3 y^{2} x + 2} + \frac{- 1}{3 y^{2} x + 2}$

Step 5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- y^{3}}{3 y^{2} x + 2} + \frac{- 1}{3 y^{2} x + 2}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

$y' = \frac{- y^{3} - 1}{3 y^{2} x + 2}$

Step 5.3.3.2

Simplify the numerator.

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Step 5.3.3.2.1

Rewrite $- y^{3}$ as ${(- y)}^{3}$ .

$y' = \frac{{(- y)}^{3} - 1}{3 y^{2} x + 2}$

Step 5.3.3.2.2

Rewrite $1$ as $1^{3}$ .

$y' = \frac{{(- y)}^{3} - 1^{3}}{3 y^{2} x + 2}$

Step 5.3.3.2.3

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = - y$ and $b = 1$ .

$y' = \frac{(- y - 1) ({(- y)}^{2} - y \cdot 1 + 1^{2})}{3 y^{2} x + 2}$

Step 5.3.3.2.4

Simplify.

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Step 5.3.3.2.4.1

Apply the product rule to $- y$ .

$y' = \frac{(- y - 1) ({(- 1)}^{2} y^{2} - y \cdot 1 + 1^{2})}{3 y^{2} x + 2}$

Step 5.3.3.2.4.2

Raise $- 1$ to the power of $2$ .

$y' = \frac{(- y - 1) (1 y^{2} - y \cdot 1 + 1^{2})}{3 y^{2} x + 2}$

Step 5.3.3.2.4.3

Multiply $y^{2}$ by $1$ .

$y' = \frac{(- y - 1) (y^{2} - y \cdot 1 + 1^{2})}{3 y^{2} x + 2}$

Step 5.3.3.2.4.4

Multiply $- 1$ by $1$ .

$y' = \frac{(- y - 1) (y^{2} - y + 1^{2})}{3 y^{2} x + 2}$

Step 5.3.3.2.4.5

One to any power is one.

$y' = \frac{(- y - 1) (y^{2} - y + 1)}{3 y^{2} x + 2}$

Step 5.3.3.3

Simplify with factoring out.

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Step 5.3.3.3.1

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

$y' = \frac{(- (y) - 1) (y^{2} - y + 1)}{3 y^{2} x + 2}$

Step 5.3.3.3.2

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

$y' = \frac{(- (y) - 1 (1)) (y^{2} - y + 1)}{3 y^{2} x + 2}$

Step 5.3.3.3.3

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

$y' = \frac{- (y + 1) (y^{2} - y + 1)}{3 y^{2} x + 2}$

Step 5.3.3.3.4

Simplify the expression.

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Step 5.3.3.3.4.1

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

$y' = \frac{- 1 (y + 1) (y^{2} - y + 1)}{3 y^{2} x + 2}$

Step 5.3.3.3.4.2

Move the negative in front of the fraction.

$y' = - \frac{(y + 1) (y^{2} - y + 1)}{3 y^{2} x + 2}$

Step 6

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

$\frac{d y}{d x} = - \frac{(y + 1) (y^{2} - y + 1)}{3 y^{2} x + 2}$