Calculus Examples

$x^{3} + 4 x y^{2} - 27 = y^{4}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{3} + 4 x y^{2} - 27) = \frac{d}{d x} (y^{4})$

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 $x^{3} + 4 x y^{2} - 27$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [4 x y^{2}] + \frac{d}{d x} [- 27]$ .

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [4 x y^{2}] + \frac{d}{d x} [- 27]$

Step 2.1.2

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

$3 x^{2} + \frac{d}{d x} [4 x y^{2}] + \frac{d}{d x} [- 27]$

Step 2.2

Evaluate $\frac{d}{d x} [4 x y^{2}]$ .

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Step 2.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x y^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x y^{2}]$ .

$3 x^{2} + 4 \frac{d}{d x} [x y^{2}] + \frac{d}{d x} [- 27]$

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

$3 x^{2} + 4 (x \frac{d}{d x} [y^{2}] + y^{2} \frac{d}{d x} [x]) + \frac{d}{d x} [- 27]$

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 $y$ .

$3 x^{2} + 4 (x (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x]) + \frac{d}{d x} [- 27]$

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 $n = 2$ .

$3 x^{2} + 4 (x (2 u \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x]) + \frac{d}{d x} [- 27]$

Step 2.2.3.3

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

$3 x^{2} + 4 (x (2 y \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x]) + \frac{d}{d x} [- 27]$

Step 2.2.4

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

$3 x^{2} + 4 (x (2 y y') + y^{2} \frac{d}{d x} [x]) + \frac{d}{d x} [- 27]$

Step 2.2.5

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

$3 x^{2} + 4 (x (2 y y') + y^{2} \cdot 1) + \frac{d}{d x} [- 27]$

Step 2.2.6

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

$3 x^{2} + 4 (2 \cdot x y y' + y^{2} \cdot 1) + \frac{d}{d x} [- 27]$

Step 2.2.7

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

$3 x^{2} + 4 (2 x y y' + y^{2}) + \frac{d}{d x} [- 27]$

Step 2.3

Since $- 27$ is constant with respect to $x$ , the derivative of $- 27$ with respect to $x$ is $0$ .

$3 x^{2} + 4 (2 x y y' + y^{2}) + 0$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$3 x^{2} + 4 (2 x y y') + 4 y^{2} + 0$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Multiply $2$ by $4$ .

$3 x^{2} + 8 (x y y') + 4 y^{2} + 0$

Step 2.4.2.2

Add $3 x^{2} + 8 x y y' + 4 y^{2}$ and $0$ .

$3 x^{2} + 8 x y y' + 4 y^{2}$

Step 2.4.3

Reorder terms.

$3 x^{2} + 4 y^{2} + 8 x y y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = y$ .

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Step 3.1.1

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

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

Step 3.1.2

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

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

Step 3.1.3

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

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

Step 3.2

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

$4 y^{3} y'$

Step 4

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

$3 x^{2} + 4 y^{2} + 8 x y y' = 4 y^{3} y'$

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $4 y^{3} y'$ from both sides of the equation.

$3 x^{2} + 4 y^{2} + 8 x y y' - 4 y^{3} y' = 0$

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 $3 x^{2}$ from both sides of the equation.

$4 y^{2} + 8 x y y' - 4 y^{3} y' = - 3 x^{2}$

Step 5.2.2

Subtract $4 y^{2}$ from both sides of the equation.

$8 x y y' - 4 y^{3} y' = - 3 x^{2} - 4 y^{2}$

Step 5.3

Factor $4 y y'$ out of $8 x y y' - 4 y^{3} y'$ .

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Step 5.3.1

Factor $4 y y'$ out of $8 x y y'$ .

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

Step 5.3.2

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

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

Step 5.3.3

Factor $4 y y'$ out of $4 y y' (2 x) + 4 y y' (- y^{2})$ .

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $4$ .

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Rewrite the expression.

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

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.

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

Step 5.4.2.2.2

Rewrite the expression.

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

Step 5.4.2.3

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

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Step 5.4.2.3.1

Cancel the common factor.

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

Step 5.4.2.3.2

Divide $y'$ by $1$ .

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

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

Move the negative in front of the fraction.

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

Step 5.4.3.1.2

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

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Step 5.4.3.1.2.1

Factor $4$ out of $- 4 y^{2}$ .

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

Step 5.4.3.1.2.2

Cancel the common factors.

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Step 5.4.3.1.2.2.1

Factor $4$ out of $4 y (2 x - y^{2})$ .

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

Step 5.4.3.1.2.2.2

Cancel the common factor.

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

Step 5.4.3.1.2.2.3

Rewrite the expression.

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

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 $- y^{2}$ .

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

Step 5.4.3.1.3.2

Cancel the common factors.

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Step 5.4.3.1.3.2.1

Cancel the common factor.

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

Step 5.4.3.1.3.2.2

Rewrite the expression.

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

Step 5.4.3.1.4

Move the negative in front of the fraction.

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

Step 5.4.3.2

To write $- \frac{y}{2 x - y^{2}}$ as a fraction with a common denominator, multiply by $\frac{4 y}{4 y}$ .

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

Step 5.4.3.3

Write each expression with a common denominator of $4 y (2 x - y^{2})$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.4.3.3.1

Multiply $\frac{y}{2 x - y^{2}}$ by $\frac{4 y}{4 y}$ .

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

Step 5.4.3.3.2

Reorder the factors of $(2 x - y^{2}) (4 y)$ .

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

Step 5.4.3.4

Combine the numerators over the common denominator.

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

Step 5.4.3.5

Simplify the numerator.

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Step 5.4.3.5.1

Rewrite using the commutative property of multiplication.

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

Step 5.4.3.5.2

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

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Step 5.4.3.5.2.1

Move $y$ .

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

Step 5.4.3.5.2.2

Multiply $y$ by $y$ .

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

Step 5.4.3.5.3

Multiply $- 1$ by $4$ .

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

Step 5.4.3.6

Simplify with factoring out.

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Step 5.4.3.6.1

Factor $- 1$ out of $- 3 x^{2}$ .

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

Step 5.4.3.6.2

Factor $- 1$ out of $- 4 y^{2}$ .

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

Step 5.4.3.6.3

Factor $- 1$ out of $- (3 x^{2}) - (4 y^{2})$ .

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

Step 5.4.3.6.4

Simplify the expression.

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Step 5.4.3.6.4.1

Rewrite $- (3 x^{2} + 4 y^{2})$ as $- 1 (3 x^{2} + 4 y^{2})$ .

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

Step 5.4.3.6.4.2

Move the negative in front of the fraction.

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

Step 6

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

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