Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

Differentiate.

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Step 2.1.1

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

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

Step 2.1.2

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

$0 + \frac{d}{d x} [- x^{2}]$

Step 2.2

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

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Step 2.2.1

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

$0 - \frac{d}{d x} [x^{2}]$

Step 2.2.2

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

$0 - (2 x)$

Step 2.2.3

Multiply $2$ by $- 1$ .

$0 - 2 x$

Step 2.3

Subtract $2 x$ from $0$ .

$- 2 x$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = y$ .

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Step 3.2.2.1

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

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

Step 3.2.2.2

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

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

Step 3.2.2.3

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

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

Step 3.2.3

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

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

Step 3.2.4

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

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

Step 3.2.5

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

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

Step 3.2.6

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

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

Step 3.3

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

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Step 3.3.1

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 3.3.1.1

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

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

Step 3.3.1.2

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

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

Step 3.3.1.3

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

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

Step 3.3.2

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

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

Step 3.4

Reorder terms.

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

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Rewrite the equation as $y^{2} + 2 x y y' + 3 y^{2} y' = - 2 x$ .

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

Step 5.2

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 5.3.3

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

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

Step 5.4

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

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Step 5.4.1

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

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

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.

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

Step 5.4.2.1.2

Rewrite the expression.

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

Step 5.4.2.2

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

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Step 5.4.2.2.1

Cancel the common factor.

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

Step 5.4.2.2.2

Divide $y'$ by $1$ .

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

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.

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

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 $- y^{2}$ .

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

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.

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

Step 5.4.3.1.2.2.2

Rewrite the expression.

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

Step 5.4.3.1.3

Move the negative in front of the fraction.

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

Step 5.4.3.2

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

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

Step 5.4.3.3

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

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Step 5.4.3.3.1

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

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

Step 5.4.3.3.2

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

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

Step 5.4.3.4

Combine the numerators over the common denominator.

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

Step 5.4.3.5

Multiply $y$ by $y$ by adding the exponents.

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Step 5.4.3.5.1

Move $y$ .

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

Step 5.4.3.5.2

Multiply $y$ by $y$ .

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

Step 5.4.3.6

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

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

Step 5.4.3.7

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

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

Step 5.4.3.8

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

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

Step 5.4.3.9

Simplify the expression.

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Step 5.4.3.9.1

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

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

Step 5.4.3.9.2

Move the negative in front of the fraction.

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

Step 6

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

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