Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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

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Step 2.2.1.1

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

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

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 $n = 3$ .

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

Step 2.2.1.3

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

$3 y^{2} y' + 2 \cdot 1$

Step 2.3.3

Multiply $2$ by $1$ .

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

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

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

Step 3.1.2

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

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

Step 3.1.3

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$2 x + 0 + \frac{d}{d x} [y^{2}]$

Step 3.2

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

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

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Step 3.2.1.1

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

$2 x + 0 + \frac{d}{d u} [u^{2}] \frac{d}{d x} [y]$

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 $n = 2$ .

$2 x + 0 + 2 u \frac{d}{d x} [y]$

Step 3.2.1.3

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

$2 x + 0 + 2 y \frac{d}{d x} [y]$

Step 3.2.2

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

$2 x + 0 + 2 y y'$

Step 3.3

Simplify.

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Step 3.3.1

Add $2 x$ and $0$ .

$2 x + 2 y y'$

Step 3.3.2

Reorder terms.

$2 y y' + 2 x$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $2 y y'$ from both sides of the equation.

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

Step 5.2

Subtract $2$ from both sides of the equation.

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

Step 5.3

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

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Step 5.3.1

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

$y y' (3 y) - 2 y y' = 2 x - 2$

Step 5.3.2

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

$y y' (3 y) + y y' \cdot - 2 = 2 x - 2$

Step 5.3.3

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

$y y' (3 y - 2) = 2 x - 2$

Step 5.4

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

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Step 5.4.1

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

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

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.

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

Step 5.4.2.1.2

Rewrite the expression.

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

Step 5.4.2.2

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

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Step 5.4.2.2.1

Cancel the common factor.

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

Step 5.4.2.2.2

Divide $y'$ by $1$ .

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

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.

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

Step 6

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

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