Calculus Examples

$0 = - 2 y^{2} + x^{3} - 5 y^{3}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (0) = \frac{d}{d x} (- 2 y^{2} + x^{3} - 5 y^{3})$

Step 2

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

$0$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $- 2 y^{2} + x^{3} - 5 y^{3}$ with respect to $x$ is $\frac{d}{d x} [- 2 y^{2}] + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 5 y^{3}]$ .

$\frac{d}{d x} [- 2 y^{2}] + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 5 y^{3}]$

Step 3.2

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

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Step 3.2.1

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

$- 2 \frac{d}{d x} [y^{2}] + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 5 y^{3}]$

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 $y$ .

$- 2 (\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [y]) + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 5 y^{3}]$

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 $n = 2$ .

$- 2 (2 u_{1} \frac{d}{d x} [y]) + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 5 y^{3}]$

Step 3.2.2.3

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

$- 2 (2 y \frac{d}{d x} [y]) + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 5 y^{3}]$

Step 3.2.3

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

$- 2 (2 y y') + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 5 y^{3}]$

Step 3.2.4

Multiply $2$ by $- 2$ .

$- 4 y y' + \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 5 y^{3}]$

Step 3.3

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

$- 4 y y' + 3 x^{2} + \frac{d}{d x} [- 5 y^{3}]$

Step 3.4

Evaluate $\frac{d}{d x} [- 5 y^{3}]$ .

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Step 3.4.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 y^{3}$ with respect to $x$ is $- 5 \frac{d}{d x} [y^{3}]$ .

$- 4 y y' + 3 x^{2} - 5 \frac{d}{d x} [y^{3}]$

Step 3.4.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = y$ .

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Step 3.4.2.1

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

$- 4 y y' + 3 x^{2} - 5 (\frac{d}{d u_{2}} [{u_{2}}^{3}] \frac{d}{d x} [y])$

Step 3.4.2.2

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

$- 4 y y' + 3 x^{2} - 5 (3 {u_{2}}^{2} \frac{d}{d x} [y])$

Step 3.4.2.3

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

$- 4 y y' + 3 x^{2} - 5 (3 y^{2} \frac{d}{d x} [y])$

Step 3.4.3

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

$- 4 y y' + 3 x^{2} - 5 (3 y^{2} y')$

Step 3.4.4

Multiply $3$ by $- 5$ .

$- 4 y y' + 3 x^{2} - 15 y^{2} y'$

Step 3.5

Reorder terms.

$3 x^{2} - 4 y y' - 15 y^{2} y'$

Step 4

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

$0 = 3 x^{2} - 4 y y' - 15 y^{2} y'$

Step 5

Solve for $y'$ .

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Step 5.1

Rewrite the equation as $3 x^{2} - 4 y y' - 15 y^{2} y' = 0$ .

$3 x^{2} - 4 y y' - 15 y^{2} y' = 0$

Step 5.2

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

$- 4 y y' - 15 y^{2} y' = - 3 x^{2}$

Step 5.3

Factor $y y'$ out of $- 4 y y' - 15 y^{2} y'$ .

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Step 5.3.1

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

$y y' \cdot - 4 - 15 y^{2} y' = - 3 x^{2}$

Step 5.3.2

Factor $y y'$ out of $- 15 y^{2} y'$ .

$y y' \cdot - 4 + y y' (- 15 y) = - 3 x^{2}$

Step 5.3.3

Factor $y y'$ out of $y y' \cdot - 4 + y y' (- 15 y)$ .

$y y' (- 4 - 15 y) = - 3 x^{2}$

Step 5.4

Divide each term in $y y' (- 4 - 15 y) = - 3 x^{2}$ by $y (- 4 - 15 y)$ and simplify.

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Step 5.4.1

Divide each term in $y y' (- 4 - 15 y) = - 3 x^{2}$ by $y (- 4 - 15 y)$ .

$\frac{y y' (- 4 - 15 y)}{y (- 4 - 15 y)} = \frac{- 3 x^{2}}{y (- 4 - 15 y)}$

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.

$\frac{y y' (- 4 - 15 y)}{y (- 4 - 15 y)} = \frac{- 3 x^{2}}{y (- 4 - 15 y)}$

Step 5.4.2.1.2

Rewrite the expression.

$\frac{y' (- 4 - 15 y)}{- 4 - 15 y} = \frac{- 3 x^{2}}{y (- 4 - 15 y)}$

Step 5.4.2.2

Cancel the common factor of $- 4 - 15 y$ .

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Step 5.4.2.2.1

Cancel the common factor.

$\frac{y' (- 4 - 15 y)}{- 4 - 15 y} = \frac{- 3 x^{2}}{y (- 4 - 15 y)}$

Step 5.4.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 3 x^{2}}{y (- 4 - 15 y)}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Move the negative in front of the fraction.

$y' = - \frac{3 x^{2}}{y (- 4 - 15 y)}$

Step 5.4.3.2

Rewrite $- 4$ as $- 1 (4)$ .

$y' = - \frac{3 x^{2}}{y (- 1 (4) - 15 y)}$

Step 5.4.3.3

Factor $- 1$ out of $- 15 y$ .

$y' = - \frac{3 x^{2}}{y (- 1 (4) - (15 y))}$

Step 5.4.3.4

Factor $- 1$ out of $- 1 (4) - (15 y)$ .

$y' = - \frac{3 x^{2}}{y (- 1 (4 + 15 y))}$

Step 5.4.3.5

Simplify the expression.

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Step 5.4.3.5.1

Move the negative in front of the fraction.

$y' = - - \frac{3 x^{2}}{y (4 + 15 y)}$

Step 5.4.3.5.2

Multiply $- 1$ by $- 1$ .

$y' = 1 \frac{3 x^{2}}{y (4 + 15 y)}$

Step 5.4.3.5.3

Multiply $\frac{3 x^{2}}{y (4 + 15 y)}$ by $1$ .

$y' = \frac{3 x^{2}}{y (4 + 15 y)}$

Step 6

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

$\frac{d y}{d x} = \frac{3 x^{2}}{y (4 + 15 y)}$