Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

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 $1 - 3 y^{2} - x y^{3}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 3 y^{2}] + \frac{d}{d x} [- x y^{3}]$ .

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

Step 2.1.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

Step 2.2

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

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Step 2.2.1

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

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

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

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Step 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $2$ by $- 3$ .

$0 - 6 y y' + \frac{d}{d x} [- x y^{3}]$

Step 2.3

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

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Step 2.3.1

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

$0 - 6 y y' - \frac{d}{d x} [x y^{3}]$

Step 2.3.2

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

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

Step 2.3.3

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 2.3.3.1

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

$0 - 6 y y' - (x (\frac{d}{d u_{2}} [{u_{2}}^{3}] \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x])$

Step 2.3.3.2

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

$0 - 6 y y' - (x (3 {u_{2}}^{2} \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x])$

Step 2.3.3.3

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

$0 - 6 y y' - (x (3 y^{2} \frac{d}{d x} [y]) + y^{3} \frac{d}{d x} [x])$

Step 2.3.4

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

$0 - 6 y y' - (x (3 y^{2} y') + y^{3} \frac{d}{d x} [x])$

Step 2.3.5

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

$0 - 6 y y' - (x (3 y^{2} y') + y^{3} \cdot 1)$

Step 2.3.6

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

$0 - 6 y y' - (3 \cdot x y^{2} y' + y^{3} \cdot 1)$

Step 2.3.7

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

$0 - 6 y y' - (3 x y^{2} y' + y^{3})$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$0 - 6 y y' - (3 x y^{2} y') - y^{3}$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Subtract $6 y y'$ from $0$ .

$- 6 y y' - (3 x y^{2} y') - y^{3}$

Step 2.4.2.2

Multiply $3$ by $- 1$ .

$- 6 y y' - 3 x y^{2} y' - y^{3}$

Step 2.4.3

Reorder terms.

$- y^{3} - 6 y y' - 3 y^{2} x y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

$- \frac{d}{d x} [x]$

Step 3.2

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

$- 1 \cdot 1$

Step 3.3

Multiply $- 1$ by $1$ .

$- 1$

Step 4

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

$- y^{3} - 6 y y' - 3 y^{2} x y' = - 1$

Step 5

Solve for $y'$ .

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Step 5.1

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

$- 6 y y' - 3 y^{2} x y' = - 1 + y^{3}$

Step 5.2

Factor $3 y y'$ out of $- 6 y y' - 3 y^{2} x y'$ .

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Step 5.2.1

Factor $3 y y'$ out of $- 6 y y'$ .

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

Step 5.2.2

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

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

Step 5.2.3

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $3$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Rewrite the expression.

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

Step 5.3.2.2

Cancel the common factor of $y$ .

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Step 5.3.2.2.1

Cancel the common factor.

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

Step 5.3.2.2.2

Rewrite the expression.

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

Step 5.3.2.3

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

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Step 5.3.2.3.1

Cancel the common factor.

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

Step 5.3.2.3.2

Divide $y'$ by $1$ .

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

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

Move the negative in front of the fraction.

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

Step 5.3.3.1.2

Cancel the common factor of $y^{3}$ and $y$ .

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Step 5.3.3.1.2.1

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

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

Step 5.3.3.1.2.2

Cancel the common factors.

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Step 5.3.3.1.2.2.1

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

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

Step 5.3.3.1.2.2.2

Cancel the common factor.

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

Step 5.3.3.1.2.2.3

Rewrite the expression.

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

Step 6

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

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