Calculus Examples

$\frac{1}{x} - \frac{1}{y} = 1$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (\frac{1}{x} - \frac{1}{y}) = \frac{d}{d x} (1)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $\frac{1}{x} - \frac{1}{y}$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{x}] + \frac{d}{d x} [- \frac{1}{y}]$ .

$\frac{d}{d x} [\frac{1}{x}] + \frac{d}{d x} [- \frac{1}{y}]$

Step 2.2

Evaluate $\frac{d}{d x} [\frac{1}{x}]$ .

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Step 2.2.1

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 2.2.2

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

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

Step 2.3

Evaluate $\frac{d}{d x} [- \frac{1}{y}]$ .

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Step 2.3.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = - 1$ and $g (x) = \frac{1}{y}$ .

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

Step 2.3.2

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = 1$ and $g (x) = y$ .

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

Step 2.3.3

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$- x^{- 2} - \frac{y \cdot 0 - 1 \cdot 1 \frac{d}{d x} [y]}{y^{2}} + \frac{1}{y} \frac{d}{d x} [- 1]$

Step 2.3.4

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

$- x^{- 2} - \frac{y \cdot 0 - 1 \cdot 1 y'}{y^{2}} + \frac{1}{y} \frac{d}{d x} [- 1]$

Step 2.3.5

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$- x^{- 2} - \frac{y \cdot 0 - 1 \cdot 1 y'}{y^{2}} + \frac{1}{y} \cdot 0$

Step 2.3.6

Multiply $y$ by $0$ .

$- x^{- 2} - \frac{0 - 1 \cdot 1 y'}{y^{2}} + \frac{1}{y} \cdot 0$

Step 2.3.7

Multiply $- 1$ by $1$ .

$- x^{- 2} - \frac{0 - y'}{y^{2}} + \frac{1}{y} \cdot 0$

Step 2.3.8

Subtract $y'$ from $0$ .

$- x^{- 2} - \frac{- y'}{y^{2}} + \frac{1}{y} \cdot 0$

Step 2.3.9

Move the negative in front of the fraction.

$- x^{- 2} - - \frac{y'}{y^{2}} + \frac{1}{y} \cdot 0$

Step 2.3.10

Multiply $- 1$ by $- 1$ .

$- x^{- 2} + 1 \frac{y'}{y^{2}} + \frac{1}{y} \cdot 0$

Step 2.3.11

Multiply $\frac{y'}{y^{2}}$ by $1$ .

$- x^{- 2} + \frac{y'}{y^{2}} + \frac{1}{y} \cdot 0$

Step 2.3.12

Multiply $\frac{1}{y}$ by $0$ .

$- x^{- 2} + \frac{y'}{y^{2}} + 0$

Step 2.3.13

Add $\frac{y'}{y^{2}}$ and $0$ .

$- x^{- 2} + \frac{y'}{y^{2}}$

Step 2.4

Simplify.

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Step 2.4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \frac{1}{x^{2}} + \frac{y'}{y^{2}}$

Step 2.4.2

Reorder terms.

$\frac{y'}{y^{2}} - \frac{1}{x^{2}}$

Step 3

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$0$

Step 4

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

$\frac{y'}{y^{2}} - \frac{1}{x^{2}} = 0$

Step 5

Solve for $y'$ .

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Step 5.1

Add $\frac{1}{x^{2}}$ to both sides of the equation.

$\frac{y'}{y^{2}} = \frac{1}{x^{2}}$

Step 5.2

Multiply both sides by $y^{2}$ .

$\frac{y'}{y^{2}} y^{2} = \frac{1}{x^{2}} y^{2}$

Step 5.3

Simplify.

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Step 5.3.1

Simplify the left side.

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Step 5.3.1.1

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

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Step 5.3.1.1.1

Cancel the common factor.

$\frac{y'}{y^{2}} y^{2} = \frac{1}{x^{2}} y^{2}$

Step 5.3.1.1.2

Rewrite the expression.

$y' = \frac{1}{x^{2}} y^{2}$

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Combine $\frac{1}{x^{2}}$ and $y^{2}$ .

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

Step 6

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

$\frac{d y}{d x} = \frac{y^{2}}{x^{2}}$