Calculus Examples

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

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{2} + y^{2} - 4) = \frac{d}{d x} (0)$

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

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

Step 2.1.2

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

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

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 $y$ .

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

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 $n = 2$ .

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

Step 2.2.1.3

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

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

Step 2.2.2

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

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

Step 2.3

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$2 x + 2 y y' + 0$

Step 2.4

Simplify.

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Step 2.4.1

Add $2 x + 2 y y'$ and $0$ .

$2 x + 2 y y'$

Step 2.4.2

Reorder terms.

$2 y y' + 2 x$

Step 3

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

$0$

Step 4

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

$2 y y' + 2 x = 0$

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $2 x$ from both sides of the equation.

$2 y y' = - 2 x$

Step 5.2

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

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Step 5.2.1

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

$\frac{2 y y'}{2 y} = \frac{- 2 x}{2 y}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $2$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{2 y y'}{2 y} = \frac{- 2 x}{2 y}$

Step 5.2.2.1.2

Rewrite the expression.

$\frac{y y'}{y} = \frac{- 2 x}{2 y}$

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.

$\frac{y y'}{y} = \frac{- 2 x}{2 y}$

Step 5.2.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 2 x}{2 y}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Cancel the common factor of $- 2$ and $2$ .

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Step 5.2.3.1.1

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

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

Step 5.2.3.1.2

Cancel the common factors.

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Step 5.2.3.1.2.1

Factor $2$ out of $2 y$ .

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

Step 5.2.3.1.2.2

Cancel the common factor.

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

Step 5.2.3.1.2.3

Rewrite the expression.

$y' = \frac{- x}{y}$

Step 5.2.3.2

Move the negative in front of the fraction.

$y' = - \frac{x}{y}$

Step 6

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

$\frac{d y}{d x} = - \frac{x}{y}$