Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

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 $x^{4} + 3 x^{3} y^{2}$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [3 x^{3} y^{2}]$ .

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

Step 2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

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

Step 2.2

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

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Step 2.2.1

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

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

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 $f (x) = x^{3}$ and $g (x) = y^{2}$ .

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

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 $y$ .

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

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 $n = 2$ .

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

Step 2.2.3.3

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

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

Step 2.2.4

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

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

Step 2.2.5

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

$4 x^{3} + 3 (x^{3} (2 y y') + y^{2} (3 x^{2}))$

Step 2.2.6

Move $2$ to the left of $x^{3}$ .

$4 x^{3} + 3 (2 \cdot x^{3} y y' + y^{2} (3 x^{2}))$

Step 2.2.7

Move $3$ to the left of $y^{2}$ .

$4 x^{3} + 3 (2 x^{3} y y' + 3 y^{2} x^{2})$

Step 2.3

Simplify.

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Step 2.3.1

Apply the distributive property.

$4 x^{3} + 3 (2 x^{3} y y') + 3 (3 y^{2} x^{2})$

Step 2.3.2

Combine terms.

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Step 2.3.2.1

Multiply $2$ by $3$ .

$4 x^{3} + 6 (x^{3} y y') + 3 (3 y^{2} x^{2})$

Step 2.3.2.2

Multiply $3$ by $3$ .

$4 x^{3} + 6 x^{3} y y' + 9 y^{2} x^{2}$

Step 2.3.3

Reorder terms.

$4 x^{3} + 6 x^{3} y y' + 9 x^{2} y^{2}$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate.

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Step 3.1.1

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

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

Step 3.1.2

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

$1 + \frac{d}{d x} [- y^{4}]$

Step 3.2

Evaluate $\frac{d}{d x} [- y^{4}]$ .

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Step 3.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- y^{4}$ with respect to $x$ is $- \frac{d}{d x} [y^{4}]$ .

$1 - \frac{d}{d x} [y^{4}]$

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 $f (x) = x^{4}$ and $g (x) = y$ .

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Step 3.2.2.1

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

$1 - (\frac{d}{d u} [u^{4}] \frac{d}{d x} [y])$

Step 3.2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 4$ .

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

Step 3.2.2.3

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

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

Step 3.2.3

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

$1 - (4 y^{3} y')$

Step 3.2.4

Multiply $4$ by $- 1$ .

$1 - 4 y^{3} y'$

Step 3.3

Reorder terms.

$- 4 y^{3} y' + 1$

Step 4

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

$4 x^{3} + 6 x^{3} y y' + 9 x^{2} y^{2} = - 4 y^{3} y' + 1$

Step 5

Solve for $y'$ .

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Step 5.1

Add $4 y^{3} y'$ to both sides of the equation.

$4 x^{3} + 6 x^{3} y y' + 9 x^{2} y^{2} + 4 y^{3} y' = 1$

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 $4 x^{3}$ from both sides of the equation.

$6 x^{3} y y' + 9 x^{2} y^{2} + 4 y^{3} y' = 1 - 4 x^{3}$

Step 5.2.2

Subtract $9 x^{2} y^{2}$ from both sides of the equation.

$6 x^{3} y y' + 4 y^{3} y' = 1 - 4 x^{3} - 9 x^{2} y^{2}$

Step 5.3

Factor $2 y y'$ out of $6 x^{3} y y' + 4 y^{3} y'$ .

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Step 5.3.1

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

$2 y y' (3 x^{3}) + 4 y^{3} y' = 1 - 4 x^{3} - 9 x^{2} y^{2}$

Step 5.3.2

Factor $2 y y'$ out of $4 y^{3} y'$ .

$2 y y' (3 x^{3}) + 2 y y' (2 y^{2}) = 1 - 4 x^{3} - 9 x^{2} y^{2}$

Step 5.3.3

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

$2 y y' (3 x^{3} + 2 y^{2}) = 1 - 4 x^{3} - 9 x^{2} y^{2}$

Step 5.4

Divide each term in $2 y y' (3 x^{3} + 2 y^{2}) = 1 - 4 x^{3} - 9 x^{2} y^{2}$ by $2 y (3 x^{3} + 2 y^{2})$ and simplify.

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Step 5.4.1

Divide each term in $2 y y' (3 x^{3} + 2 y^{2}) = 1 - 4 x^{3} - 9 x^{2} y^{2}$ by $2 y (3 x^{3} + 2 y^{2})$ .

$\frac{2 y y' (3 x^{3} + 2 y^{2})}{2 y (3 x^{3} + 2 y^{2})} = \frac{1}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 4 x^{3}}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 9 x^{2} y^{2}}{2 y (3 x^{3} + 2 y^{2})}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $2$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{2 y y' (3 x^{3} + 2 y^{2})}{2 y (3 x^{3} + 2 y^{2})} = \frac{1}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 4 x^{3}}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 9 x^{2} y^{2}}{2 y (3 x^{3} + 2 y^{2})}$

Step 5.4.2.1.2

Rewrite the expression.

$\frac{y y' (3 x^{3} + 2 y^{2})}{y (3 x^{3} + 2 y^{2})} = \frac{1}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 4 x^{3}}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 9 x^{2} y^{2}}{2 y (3 x^{3} + 2 y^{2})}$

Step 5.4.2.2

Cancel the common factor of $y$ .

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Step 5.4.2.2.1

Cancel the common factor.

$\frac{y y' (3 x^{3} + 2 y^{2})}{y (3 x^{3} + 2 y^{2})} = \frac{1}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 4 x^{3}}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 9 x^{2} y^{2}}{2 y (3 x^{3} + 2 y^{2})}$

Step 5.4.2.2.2

Rewrite the expression.

$\frac{y' (3 x^{3} + 2 y^{2})}{3 x^{3} + 2 y^{2}} = \frac{1}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 4 x^{3}}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 9 x^{2} y^{2}}{2 y (3 x^{3} + 2 y^{2})}$

Step 5.4.2.3

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

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Step 5.4.2.3.1

Cancel the common factor.

$\frac{y' (3 x^{3} + 2 y^{2})}{3 x^{3} + 2 y^{2}} = \frac{1}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 4 x^{3}}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 9 x^{2} y^{2}}{2 y (3 x^{3} + 2 y^{2})}$

Step 5.4.2.3.2

Divide $y'$ by $1$ .

$y' = \frac{1}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 4 x^{3}}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 9 x^{2} y^{2}}{2 y (3 x^{3} + 2 y^{2})}$

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

Cancel the common factor of $- 4$ and $2$ .

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Step 5.4.3.1.1.1

Factor $2$ out of $- 4 x^{3}$ .

$y' = \frac{1}{2 y (3 x^{3} + 2 y^{2})} + \frac{2 (- 2 x^{3})}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 9 x^{2} y^{2}}{2 y (3 x^{3} + 2 y^{2})}$

Step 5.4.3.1.1.2

Cancel the common factors.

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Step 5.4.3.1.1.2.1

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

$y' = \frac{1}{2 y (3 x^{3} + 2 y^{2})} + \frac{2 (- 2 x^{3})}{2 (y (3 x^{3} + 2 y^{2}))} + \frac{- 9 x^{2} y^{2}}{2 y (3 x^{3} + 2 y^{2})}$

Step 5.4.3.1.1.2.2

Cancel the common factor.

$y' = \frac{1}{2 y (3 x^{3} + 2 y^{2})} + \frac{2 (- 2 x^{3})}{2 (y (3 x^{3} + 2 y^{2}))} + \frac{- 9 x^{2} y^{2}}{2 y (3 x^{3} + 2 y^{2})}$

Step 5.4.3.1.1.2.3

Rewrite the expression.

$y' = \frac{1}{2 y (3 x^{3} + 2 y^{2})} + \frac{- 2 x^{3}}{y (3 x^{3} + 2 y^{2})} + \frac{- 9 x^{2} y^{2}}{2 y (3 x^{3} + 2 y^{2})}$

Step 5.4.3.1.2

Move the negative in front of the fraction.

$y' = \frac{1}{2 y (3 x^{3} + 2 y^{2})} - \frac{2 x^{3}}{y (3 x^{3} + 2 y^{2})} + \frac{- 9 x^{2} y^{2}}{2 y (3 x^{3} + 2 y^{2})}$

Step 5.4.3.1.3

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

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Step 5.4.3.1.3.1

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

$y' = \frac{1}{2 y (3 x^{3} + 2 y^{2})} - \frac{2 x^{3}}{y (3 x^{3} + 2 y^{2})} + \frac{y (- 9 x^{2} y)}{2 y (3 x^{3} + 2 y^{2})}$

Step 5.4.3.1.3.2

Cancel the common factors.

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Step 5.4.3.1.3.2.1

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

$y' = \frac{1}{2 y (3 x^{3} + 2 y^{2})} - \frac{2 x^{3}}{y (3 x^{3} + 2 y^{2})} + \frac{y (- 9 x^{2} y)}{y (2 (3 x^{3} + 2 y^{2}))}$

Step 5.4.3.1.3.2.2

Cancel the common factor.

$y' = \frac{1}{2 y (3 x^{3} + 2 y^{2})} - \frac{2 x^{3}}{y (3 x^{3} + 2 y^{2})} + \frac{y (- 9 x^{2} y)}{y (2 (3 x^{3} + 2 y^{2}))}$

Step 5.4.3.1.3.2.3

Rewrite the expression.

$y' = \frac{1}{2 y (3 x^{3} + 2 y^{2})} - \frac{2 x^{3}}{y (3 x^{3} + 2 y^{2})} + \frac{- 9 x^{2} y}{2 (3 x^{3} + 2 y^{2})}$

Step 5.4.3.1.4

Move the negative in front of the fraction.

$y' = \frac{1}{2 y (3 x^{3} + 2 y^{2})} - \frac{2 x^{3}}{y (3 x^{3} + 2 y^{2})} - \frac{9 x^{2} y}{2 (3 x^{3} + 2 y^{2})}$

Step 6

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

$\frac{d y}{d x} = \frac{1}{2 y (3 x^{3} + 2 y^{2})} - \frac{2 x^{3}}{y (3 x^{3} + 2 y^{2})} - \frac{9 x^{2} y}{2 (3 x^{3} + 2 y^{2})}$