Find dy/dx square root of x+ square root of y=1

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

Step 1

Rewrite the left side with rational exponents.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

$x^{\frac{1}{2}} + \sqrt{y} = 1$

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y}$ as $y^{\frac{1}{2}}$ .

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

Step 2

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{\frac{1}{2}} + y^{\frac{1}{2}}) = \frac{d}{d x} (1)$

Step 3

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $x^{\frac{1}{2}} + y^{\frac{1}{2}}$ with respect to $x$ is $\frac{d}{d x} [x^{\frac{1}{2}}] + \frac{d}{d x} [y^{\frac{1}{2}}]$ .

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

Evaluate $\frac{d}{d x} [x^{\frac{1}{2}}]$ .

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

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Move the negative in front of the fraction.

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

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

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

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

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

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

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

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

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

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

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $1$ .

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

Move the negative in front of the fraction.

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

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

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

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

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

Move $y^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Simplify.

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

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

Multiply $\frac{1}{2}$ by $\frac{1}{x^{\frac{1}{2}}}$ .

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

Step 4

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

$0$

Step 5

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

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

Step 6

Solve for $y'$ .

Subtract $\frac{1}{2 x^{\frac{1}{2}}}$ from both sides of the equation.

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

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

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

Simplify.

Simplify the left side.

Simplify $\frac{y'}{2 y^{\frac{1}{2}}} (2 y^{\frac{1}{2}})$ .

Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $y^{\frac{1}{2}}$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Simplify $- \frac{1}{2 x^{\frac{1}{2}}} (2 y^{\frac{1}{2}})$ .

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2 x^{\frac{1}{2}}}$ into the numerator.

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

Factor $2$ out of $2 x^{\frac{1}{2}}$ .

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

Factor $2$ out of $2 y^{\frac{1}{2}}$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Move the negative in front of the fraction.

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

Step 7

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

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

Find dy/dx square root of x+ square root of y=1

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