Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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

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Step 2.2.1.1

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

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

Step 2.2.1.2

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

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

Step 2.2.1.3

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

$2 y y' + 3 (2 x)$

Step 2.3.3

Multiply $2$ by $3$ .

$2 y y' + 6 x$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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Step 3.2.1.1

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

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

Step 3.2.1.2

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

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

Step 3.2.1.3

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

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

Step 3.2.2

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

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

Multiply $3$ by $- 1$ .

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

Step 3.4

Reorder terms.

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

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

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

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

Step 5.2

Subtract $6 x$ from both sides of the equation.

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

Step 5.3

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

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Step 5.3.1

Factor $y y'$ out of $2 y y'$ .

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

Step 5.3.2

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

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

Step 5.3.3

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

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

Step 5.4

Divide each term in $y y' (2 - 3 y) = - 3 x^{2} - 6 x$ by $y (2 - 3 y)$ and simplify.

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Step 5.4.1

Divide each term in $y y' (2 - 3 y) = - 3 x^{2} - 6 x$ by $y (2 - 3 y)$ .

$\frac{y y' (2 - 3 y)}{y (2 - 3 y)} = \frac{- 3 x^{2}}{y (2 - 3 y)} + \frac{- 6 x}{y (2 - 3 y)}$

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.

$\frac{y y' (2 - 3 y)}{y (2 - 3 y)} = \frac{- 3 x^{2}}{y (2 - 3 y)} + \frac{- 6 x}{y (2 - 3 y)}$

Step 5.4.2.1.2

Rewrite the expression.

$\frac{y' (2 - 3 y)}{2 - 3 y} = \frac{- 3 x^{2}}{y (2 - 3 y)} + \frac{- 6 x}{y (2 - 3 y)}$

Step 5.4.2.2

Cancel the common factor of $2 - 3 y$ .

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Step 5.4.2.2.1

Cancel the common factor.

$\frac{y' (2 - 3 y)}{2 - 3 y} = \frac{- 3 x^{2}}{y (2 - 3 y)} + \frac{- 6 x}{y (2 - 3 y)}$

Step 5.4.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 3 x^{2}}{y (2 - 3 y)} + \frac{- 6 x}{y (2 - 3 y)}$

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.

$y' = - \frac{3 x^{2}}{y (2 - 3 y)} + \frac{- 6 x}{y (2 - 3 y)}$

Step 5.4.3.1.2

Move the negative in front of the fraction.

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

Step 6

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

$\frac{d y}{d x} = - \frac{3 x^{2}}{y (2 - 3 y)} - \frac{6 x}{y (2 - 3 y)}$