Question

find dy/dx if x + √xy + y = 1​

Answer 1

Answer:

Given x+√xy +y=1

Differentiate w.r.t.x

1+1/2√xy (x dy/dx +y 1)+dy/dx =0

Or (1+y/2√xy)+dy/dx (1+x/2√xy) =0

Or dy/dx =-(1+y/2√xy)/(1+x/2√xy)

Or dy/dx =-(1+√y/2√x)/(1+√x/2√y)

Or dy/dx =-(2√x+√y)(2√y+√x) /4√xy

Step-by-step explanation:

Answer 2

Answer:

The value of given differential is

$\frac{dy}{dx} = \frac{ - (y + 2 \sqrt{xy}) }{x + 2 \sqrt{xy} }$

Step-by-step explanation:

Given expression is

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

Differentiating with respect to x on both sides,we get

$\frac{d}{dx} (x + \sqrt{xy} + y) = 0 \\ = > 1 + \sqrt{x} \frac{d}{dx} ( \sqrt{y} ) + \sqrt{y} \frac{d}{dx} ( \sqrt{x} ) + \frac{dy}{dx} = 0 \\ = > 1 + \frac{ \sqrt{x} }{2 \sqrt{y} }( \frac{dy}{dx} ) + \frac{ \sqrt{y} }{2 \sqrt{x} } + \frac{dy}{dx} = 0 \\ = > 1 + \frac{dy}{dx} ( \frac{ \sqrt{x} }{2 \sqrt{y} } + 1) = - \frac{ \sqrt{y} }{2 \sqrt{x} } \\ = > \frac{dy}{dx} = \frac{ \frac{ - \sqrt{y} }{2 \sqrt{x} } - 1 }{ \frac{ \sqrt{x} }{2 \sqrt{y} } + 1 } = \frac{ - \sqrt{y} - 2 \sqrt{x} }{ \sqrt{x} + 2 \sqrt{y} } \times \sqrt{ \frac{y}{x} } \\ = \frac{ - (y + 2 \sqrt{xy}) }{x + 2 \sqrt{xy} }$

Therefore,the value of given differential is

$\frac{dy}{dx} = \frac{ - (y + 2 \sqrt{xy}) }{x + 2 \sqrt{xy} }$

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