Given-y-x-x-y-2-find-the-value-of-dy-dx-1-1-

Question Number 99960 by bemath last updated on 24/Jun/20

Given y \sqrt{x} + x \sqrt{y} = 2 . find the value of

\frac{dy}{dx} ∣_{(1, 1)} = ?

Commented by Dwaipayan Shikari last updated on 24/Jun/20

- 1

Answered by Rio Michael last updated on 24/Jun/20

\sqrt{x} \frac{d y}{d x} + \frac{y}{2 \sqrt{x}} + \frac{x}{2 \sqrt{y}} \frac{d y}{d x} + \sqrt{y} = 0

\Rightarrow \sqrt{1} \frac{d y}{d x} ∣_{(1, 1)} + \frac{1}{2 \sqrt{1}} + \frac{1}{2 \sqrt{1}} \frac{d y}{d x} ∣_{(1, 1)} + \sqrt{1} = 0

Answered by 1549442205 last updated on 24/Jun/20

Derivative two sides of the givenwewe equation

by x we obtain y^{^{'}} \sqrt{x} + \frac{y}{2 \sqrt{x}} + \sqrt{y} + \frac{{xy}^{'}}{2 \sqrt{y}} = 0

\Rightarrow y^{'} (\sqrt{x} + \frac{x}{2 \sqrt{y}}) = - (\frac{y}{2 \sqrt{x}} + \sqrt{y}) \Rightarrow y^{'} = \frac{- (2 \sqrt{xy} + y)}{2 \sqrt{xy} + x}

\Rightarrow y^{'} ∣_{(1, 1)} = \frac{- (2 \sqrt{xy} + y)}{2 \sqrt{xy} + x} ∣_{(1, 1)} = \frac{- 3}{3} = - 1