Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

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

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

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

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

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 $u$ as $y$ .

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

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 $n = 2$ .

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

Step 2.2.3.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

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

Step 2.2.6

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

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

Step 2.2.7

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

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

Step 2.3

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

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

Step 2.4

Evaluate $\frac{d}{d x} [2 x^{3}]$ .

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Step 2.4.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x^{3}$ with respect to $x$ is $2 \frac{d}{d x} [x^{3}]$ .

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

Step 2.4.2

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

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

Step 2.4.3

Multiply $3$ by $2$ .

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

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

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

Step 2.5.2

Combine terms.

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Step 2.5.2.1

Multiply $2$ by $- 1$ .

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

Step 2.5.2.2

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

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

Step 2.5.3

Reorder terms.

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.3

Multiply $3$ by $- 3$ .

$- 9 (y^{2} \frac{d}{d x} [y])$

Step 3.4

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

$- 9 y^{2} y'$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Add $9 y^{2} y'$ to both sides of the equation.

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

Step 5.2

Move all terms not containing $y'$ to the right side of the equation.

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Step 5.2.1

Subtract $6 x^{2}$ from both sides of the equation.

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

Step 5.2.2

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 5.3.3

Factor $y y'$ out of $y y' (- 2 x) + y y' (9 y)$ .

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $y$ .

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Rewrite the expression.

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

Step 5.4.2.2

Cancel the common factor of $- 2 x + 9 y$ .

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Step 5.4.2.2.1

Cancel the common factor.

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

Step 5.4.2.2.2

Divide $y'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Simplify each term.

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Step 5.4.3.1.1

Move the negative in front of the fraction.

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

Step 5.4.3.1.2

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

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Step 5.4.3.1.2.1

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

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

Step 5.4.3.1.2.2

Cancel the common factors.

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Step 5.4.3.1.2.2.1

Cancel the common factor.

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

Step 5.4.3.1.2.2.2

Rewrite the expression.

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

Step 5.4.3.2

To write $\frac{y}{- 2 x + 9 y}$ as a fraction with a common denominator, multiply by $\frac{y}{y}$ .

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

Step 5.4.3.3

Write each expression with a common denominator of $y (- 2 x + 9 y)$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.4.3.3.1

Multiply $\frac{y}{- 2 x + 9 y}$ by $\frac{y}{y}$ .

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

Step 5.4.3.3.2

Reorder the factors of $(- 2 x + 9 y) y$ .

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

Step 5.4.3.4

Combine the numerators over the common denominator.

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

Step 5.4.3.5

Multiply $y$ by $y$ .

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

Step 5.4.3.6

Factor $- 1$ out of $- 6 x^{2}$ .

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

Step 5.4.3.7

Factor $- 1$ out of $y^{2}$ .

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

Step 5.4.3.8

Factor $- 1$ out of $- (6 x^{2}) - 1 (- y^{2})$ .

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

Step 5.4.3.9

Rewrite $- (6 x^{2} - y^{2})$ as $- 1 (6 x^{2} - y^{2})$ .

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

Step 5.4.3.10

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

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

Step 5.4.3.11

Factor $- 1$ out of $9 y$ .

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

Step 5.4.3.12

Factor $- 1$ out of $- (2 x) - (- 9 y)$ .

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

Step 5.4.3.13

Rewrite $- (2 x - 9 y)$ as $- 1 (2 x - 9 y)$ .

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

Step 5.4.3.14

Cancel the common factor.

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

Step 5.4.3.15

Rewrite the expression.

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

Step 6

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

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