Question

for the curve √x + √y = 1 , DY/ DX at (1/9,1/9) is. plz answer​

Answer 1

Given :

• The given equation is $\sf{\sqrt{x}+\sqrt{y}=1}$

•Coordinates =$\sf{\bigg(\dfrac{1}{9},\dfrac{1}{9} \bigg) }$

To Find :

•The value of $\sf{\bigg(\dfrac{1}{9},\dfrac{1}{9} \bigg)=? }$

Solution :

•The Given equation is

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

•The Given points is >> $\sf{\bigg(\dfrac{1}{9},\dfrac{1}{9} \bigg) }$

Differentiating both sides with respect to x.

$\\ \implies \sf \: \sqrt{x} + \sqrt{y} = 1 \\ \\ \implies \sf \: \frac{1}{2 \sqrt{x} } + \frac{1}{2 \sqrt{y} } . \frac{dy}{dx} = 0 \\ \\ \implies \sf \: \frac{1}{2 \sqrt{y} } . \frac{dy}{dx} = - \frac{1}{2 \sqrt{x} } \\ \\ \implies \sf \: \frac{dy}{dx} = - \frac{2 \sqrt{y} }{2 \sqrt{x} } \\ \\ \implies \sf \frac{dy}{dx} = \frac{ - \sqrt{y} }{ \sqrt{x} }$

Substituting the coordinates in given equation :

$\\ \implies\sf{\bigg(\dfrac{dy}{dx} \bigg)=-\dfrac{\sqrt{y}}{\sqrt{x}} }$

$\\ \implies \sf \: \frac{dy}{dx} = - \frac{ \sqrt{ \frac{1}{9} } }{ \sqrt{ \frac{1}{9} } } \\ \\ \implies \sf \: \frac{dy}{dx} = - 1 \\ \\ \implies \boxed{ \sf{ \frac{dy}{dx} = - 1}}$

• The value of dy/dx is -1