Find dy/dx 1/x+1/y=1

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

Differentiate both sides of the equation.

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

Differentiate the left side of the equation.

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

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

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

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

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

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

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

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

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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 $g (x) = y$ .

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

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

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

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

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

Multiply $y$ by $0$ .

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

Multiply $- 1$ by $1$ .

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

Subtract $\frac{d}{d x} [y]$ from $0$ .

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

Move the negative in front of the fraction.

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

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

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

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

$0$

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

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

Solve for $y'$ .

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

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

Multiply each term in $- \frac{y'}{y^{2}} = \frac{1}{x^{2}}$ by $- 1$

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Multiply each term in $- \frac{y'}{y^{2}} = \frac{1}{x^{2}}$ by $- 1$ .

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

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

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Multiply $- 1$ by $- 1$ .

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

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

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

Simplify $\frac{1}{x^{2}} \cdot - 1$ .

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

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

Move the negative in front of the fraction.

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

Multiply both sides of the equation by $y^{2}$ .

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

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

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

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

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

Find dy/dx 1/x+1/y=1

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