If x-y-y-x-1- then dy-dx equals

The correct option is D √(y/x) ( y+2√(xy)/x+2√(xy) ) x√( y ) +y√( x ) =1 #160;⇒√( y ) + x / 2√( y ) #160; dy / dx + dy / dx √( x ) +y 1 / 2√ ...

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Standard XII

Mathematics

Implicit Differentiation

If x√y+y√x=...

Question

If $x \sqrt{y} + y \sqrt{x} = 1$ , then $\frac{d y}{d x}$ equals

- \frac{y + 2 \sqrt{x y}}{x + 2 \sqrt{x y}}

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- \sqrt{\frac{x}{y}} (\frac{y + 2 \sqrt{x y}}{x + 2 \sqrt{x y}})

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- \sqrt{\frac{y}{x}} (\frac{y + 2 \sqrt{x y}}{x + 2 \sqrt{x y}})

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Solution

The correct option is D $- \sqrt{\frac{y}{x}} (\frac{y + 2 \sqrt{x y}}{x + 2 \sqrt{x y}})$
$x \sqrt{y} + y \sqrt{x} = 1$
$\Rightarrow \sqrt{y} + \frac{x}{2 \sqrt{y}} \frac{d y}{d x} + \frac{d y}{d x} \sqrt{x} + y \frac{1}{2 \sqrt{x}} = 0$
$\Rightarrow (\frac{x}{2 \sqrt{y}} + \sqrt{x}) \frac{d y}{d x} = - \frac{y}{2 \sqrt{x}} - \sqrt{y}$
$\Rightarrow (\frac{x + 2 \sqrt{x y}}{2 \sqrt{y}}) \frac{d y}{d x} = - \frac{y - 2 \sqrt{x y}}{2 \sqrt{x}}$
$\Rightarrow \frac{d y}{d x} = - \sqrt{\frac{y}{x}} (\frac{y + 2 \sqrt{x y}}{x + 2 \sqrt{x y}})$

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Similar questions

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

Which of the following differential equations has y=x as one of its particular solution?
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If x√y+y√x=1, then dydx equals

The correct option is D −√yx(y+2√xyx+2√xy)x√y+y√x=1⇒√y+x2√ydydx+dydx√x+y12√x=0⇒(x2√y+√x)dydx=−y2√x−√y⇒(x+2√xy2√y)dydx=−y−2√xy2√x⇒dydx=−√yx(y+2√xyx+2√xy)

If $x \sqrt{y} + y \sqrt{x} = 1$ , then $\frac{d y}{d x}$ equals