Calculus Examples

$- x y^{2} + x + 4 - y^{3} = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (- x y^{2} + x + 4 - y^{3}) = \frac{d}{d x} (0)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $- x y^{2} + x + 4 - y^{3}$ with respect to $x$ is $\frac{d}{d x} [- x y^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [4] + \frac{d}{d x} [- y^{3}]$ .

$\frac{d}{d x} [- x y^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [4] + \frac{d}{d x} [- y^{3}]$

Step 2.2

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

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Step 2.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x y^{2}$ with respect to $x$ is $- \frac{d}{d x} [x y^{2}]$ .

$- \frac{d}{d x} [x y^{2}] + \frac{d}{d x} [x] + \frac{d}{d x} [4] + \frac{d}{d x} [- y^{3}]$

Step 2.2.2

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 $g (x) = y^{2}$ .

$- (x \frac{d}{d x} [y^{2}] + y^{2} \frac{d}{d x} [x]) + \frac{d}{d x} [x] + \frac{d}{d x} [4] + \frac{d}{d x} [- y^{3}]$

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 $f (x) = x^{2}$ and $g (x) = y$ .

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Step 2.2.3.1

To apply the Chain Rule, set $u_{1}$ as $y$ .

$- (x (\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x]) + \frac{d}{d x} [x] + \frac{d}{d x} [4] + \frac{d}{d x} [- y^{3}]$

Step 2.2.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 2$ .

$- (x (2 u_{1} \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x]) + \frac{d}{d x} [x] + \frac{d}{d x} [4] + \frac{d}{d x} [- y^{3}]$

Step 2.2.3.3

Replace all occurrences of $u_{1}$ with $y$ .

$- (x (2 y \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x]) + \frac{d}{d x} [x] + \frac{d}{d x} [4] + \frac{d}{d x} [- y^{3}]$

Step 2.2.4

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

$- (x (2 y y') + y^{2} \frac{d}{d x} [x]) + \frac{d}{d x} [x] + \frac{d}{d x} [4] + \frac{d}{d x} [- y^{3}]$

Step 2.2.5

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

$- (x (2 y y') + y^{2} \cdot 1) + \frac{d}{d x} [x] + \frac{d}{d x} [4] + \frac{d}{d x} [- y^{3}]$

Step 2.2.6

Move $2$ to the left of $x$ .

$- (2 \cdot x y y' + y^{2} \cdot 1) + \frac{d}{d x} [x] + \frac{d}{d x} [4] + \frac{d}{d x} [- y^{3}]$

Step 2.2.7

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

$- (2 x y y' + y^{2}) + \frac{d}{d x} [x] + \frac{d}{d x} [4] + \frac{d}{d x} [- y^{3}]$

Step 2.3

Differentiate.

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Step 2.3.1

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

$- (2 x y y' + y^{2}) + 1 + \frac{d}{d x} [4] + \frac{d}{d x} [- y^{3}]$

Step 2.3.2

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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 2.4.2.1

To apply the Chain Rule, set $u_{2}$ as $y$ .

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

Step 2.4.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 3$ .

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

Step 2.4.2.3

Replace all occurrences of $u_{2}$ with $y$ .

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

Step 2.4.3

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

$- (2 x y y' + y^{2}) + 1 + 0 - (3 y^{2} y')$

Step 2.4.4

Multiply $3$ by $- 1$ .

$- (2 x y y' + y^{2}) + 1 + 0 - 3 y^{2} y'$

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

$- (2 x y y') - y^{2} + 1 + 0 - 3 y^{2} y'$

Step 2.5.2

Combine terms.

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Step 2.5.2.1

Multiply $2$ by $- 1$ .

$- 2 (x y y') - y^{2} + 1 + 0 - 3 y^{2} y'$

Step 2.5.2.2

Add $- 2 x y y' - y^{2} + 1$ and $0$ .

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

Step 2.5.3

Reorder terms.

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

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.

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

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

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

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

Step 5.1.2

Subtract $1$ from both sides of the equation.

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.2.3

Factor $y y'$ out of $y y' (- 2 x) + y y' (- 3 y)$ .

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $y$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Rewrite the expression.

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

Step 5.3.2.2

Cancel the common factor of $- 2 x - 3 y$ .

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Step 5.3.2.2.1

Cancel the common factor.

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

Step 5.3.2.2.2

Divide $y'$ by $1$ .

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Simplify each term.

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Step 5.3.3.1.1

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

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Step 5.3.3.1.1.1

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

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

Step 5.3.3.1.1.2

Cancel the common factors.

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Step 5.3.3.1.1.2.1

Cancel the common factor.

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

Step 5.3.3.1.1.2.2

Rewrite the expression.

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

Step 5.3.3.1.2

Move the negative in front of the fraction.

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

Step 5.3.3.2

To write $\frac{y}{- 2 x - 3 y}$ as a fraction with a common denominator, multiply by $\frac{y}{y}$ .

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

Step 5.3.3.3

Write each expression with a common denominator of $(- 2 x - 3 y) y$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.3.3.3.1

Multiply $\frac{y}{- 2 x - 3 y}$ by $\frac{y}{y}$ .

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

Step 5.3.3.3.2

Reorder the factors of $(- 2 x - 3 y) y$ .

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

Step 5.3.3.4

Combine the numerators over the common denominator.

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

Step 5.3.3.5

Simplify the numerator.

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Step 5.3.3.5.1

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

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

Step 5.3.3.5.2

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

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

Step 5.3.3.5.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 5.3.3.5.4

Add $1$ and $1$ .

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

Step 5.3.3.5.5

Rewrite $1$ as $1^{2}$ .

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

Step 5.3.3.5.6

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

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

Step 5.3.3.6

Simplify with factoring out.

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Step 5.3.3.6.1

Factor $- 1$ out of $- 2 x$ .

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

Step 5.3.3.6.2

Factor $- 1$ out of $- 3 y$ .

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

Step 5.3.3.6.3

Factor $- 1$ out of $- (2 x) - (3 y)$ .

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

Step 5.3.3.6.4

Rewrite negatives.

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Step 5.3.3.6.4.1

Rewrite $- (2 x + 3 y)$ as $- 1 (2 x + 3 y)$ .

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

Step 5.3.3.6.4.2

Move the negative in front of the fraction.

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

Step 6

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

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