Calculus Examples

$y^{4} - 4 y^{2} = x^{4} - 9 x^{2}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y^{4} - 4 y^{2}) = \frac{d}{d x} (x^{4} - 9 x^{2})$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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 2.2.1.1

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

$\frac{d}{d u_{1}} [{u_{1}}^{4}] \frac{d}{d x} [y] + \frac{d}{d x} [- 4 y^{2}]$

Step 2.2.1.2

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

$4 {u_{1}}^{3} \frac{d}{d x} [y] + \frac{d}{d x} [- 4 y^{2}]$

Step 2.2.1.3

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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Step 2.3.2.1

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

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

Step 2.3.2.2

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

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

Step 2.3.2.3

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

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

Step 2.3.3

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

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

Step 2.3.4

Multiply $2$ by $- 4$ .

$4 y^{3} y' - 8 y y'$

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 $x^{4} - 9 x^{2}$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 9 x^{2}]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 9 x^{2}]$

Step 3.1.2

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

$4 x^{3} + \frac{d}{d x} [- 9 x^{2}]$

Step 3.2

Evaluate $\frac{d}{d x} [- 9 x^{2}]$ .

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Step 3.2.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 x^{2}$ with respect to $x$ is $- 9 \frac{d}{d x} [x^{2}]$ .

$4 x^{3} - 9 \frac{d}{d x} [x^{2}]$

Step 3.2.2

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

$4 x^{3} - 9 (2 x)$

Step 3.2.3

Multiply $2$ by $- 9$ .

$4 x^{3} - 18 x$

Step 4

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

$4 y^{3} y' - 8 y y' = 4 x^{3} - 18 x$

Step 5

Solve for $y'$ .

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Step 5.1

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

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Step 5.1.1

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

$4 y y' y^{2} - 8 y y' = 4 x^{3} - 18 x$

Step 5.1.2

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

$4 y y' y^{2} + 4 y y' \cdot - 2 = 4 x^{3} - 18 x$

Step 5.1.3

Factor $4 y y'$ out of $4 y y' y^{2} + 4 y y' \cdot - 2$ .

$4 y y' (y^{2} - 2) = 4 x^{3} - 18 x$

Step 5.2

Divide each term in $4 y y' (y^{2} - 2) = 4 x^{3} - 18 x$ by $4 y (y^{2} - 2)$ and simplify.

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Step 5.2.1

Divide each term in $4 y y' (y^{2} - 2) = 4 x^{3} - 18 x$ by $4 y (y^{2} - 2)$ .

$\frac{4 y y' (y^{2} - 2)}{4 y (y^{2} - 2)} = \frac{4 x^{3}}{4 y (y^{2} - 2)} + \frac{- 18 x}{4 y (y^{2} - 2)}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $4$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{4 y y' (y^{2} - 2)}{4 y (y^{2} - 2)} = \frac{4 x^{3}}{4 y (y^{2} - 2)} + \frac{- 18 x}{4 y (y^{2} - 2)}$

Step 5.2.2.1.2

Rewrite the expression.

$\frac{y y' (y^{2} - 2)}{y (y^{2} - 2)} = \frac{4 x^{3}}{4 y (y^{2} - 2)} + \frac{- 18 x}{4 y (y^{2} - 2)}$

Step 5.2.2.2

Cancel the common factor of $y$ .

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Step 5.2.2.2.1

Cancel the common factor.

$\frac{y y' (y^{2} - 2)}{y (y^{2} - 2)} = \frac{4 x^{3}}{4 y (y^{2} - 2)} + \frac{- 18 x}{4 y (y^{2} - 2)}$

Step 5.2.2.2.2

Rewrite the expression.

$\frac{y' (y^{2} - 2)}{y^{2} - 2} = \frac{4 x^{3}}{4 y (y^{2} - 2)} + \frac{- 18 x}{4 y (y^{2} - 2)}$

Step 5.2.2.3

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

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Step 5.2.2.3.1

Cancel the common factor.

$\frac{y' (y^{2} - 2)}{y^{2} - 2} = \frac{4 x^{3}}{4 y (y^{2} - 2)} + \frac{- 18 x}{4 y (y^{2} - 2)}$

Step 5.2.2.3.2

Divide $y'$ by $1$ .

$y' = \frac{4 x^{3}}{4 y (y^{2} - 2)} + \frac{- 18 x}{4 y (y^{2} - 2)}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Simplify each term.

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Step 5.2.3.1.1

Cancel the common factor of $4$ .

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Step 5.2.3.1.1.1

Cancel the common factor.

$y' = \frac{4 x^{3}}{4 y (y^{2} - 2)} + \frac{- 18 x}{4 y (y^{2} - 2)}$

Step 5.2.3.1.1.2

Rewrite the expression.

$y' = \frac{x^{3}}{y (y^{2} - 2)} + \frac{- 18 x}{4 y (y^{2} - 2)}$

Step 5.2.3.1.2

Cancel the common factor of $- 18$ and $4$ .

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Step 5.2.3.1.2.1

Factor $2$ out of $- 18 x$ .

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

Step 5.2.3.1.2.2

Cancel the common factors.

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Step 5.2.3.1.2.2.1

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

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

Step 5.2.3.1.2.2.2

Cancel the common factor.

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

Step 5.2.3.1.2.2.3

Rewrite the expression.

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

Step 5.2.3.1.3

Move the negative in front of the fraction.

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

Step 6

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

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