Question

Find dy/DX if√x+√y=√a

Answer 1

1/2rootx+1/2rooty.dy/dx=0
=》1/2rooty. dy/dx= -1/2rootx
=》dy/dx= rooty/rootx


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Answer 2

$\displaystyle \sf{If \: \: \sqrt{x} + \sqrt{y} = \sqrt{a} \: \: then \: \: \frac{dy}{dx} = 1 - \sqrt{ \frac{a}{x} } }$

Given :

$\displaystyle \sf{\sqrt{x} + \sqrt{y} = \sqrt{a} }$

To find :

$\displaystyle \sf \: \frac{dy}{dx}$

Solution :

Step 1 of 3 :

Write down the given equation

The given equation is

$\displaystyle \sf{ \sqrt{x} + \sqrt{y} = \sqrt{a} }$

Step 2 of 3 :

Find the value of y

$\displaystyle \sf{ \sqrt{x} + \sqrt{y} = \sqrt{a} }$

$\displaystyle \sf{ \sqrt{y} = \sqrt{a} - \sqrt{x} }$

Squaring both sides we get

$\displaystyle \sf{ y = {( \sqrt{a} - \sqrt{x} )}^{2} }$

$\displaystyle \sf{ \implies y = a + x - 2 \sqrt{ax} }$

Step 3 of 3 :

Differentiate both sides

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \implies \frac{dy}{dx} = 0 + 1 - 2 \sqrt{a} \times \frac{1}{2} {x}^{ \frac{1}{2} - 1 } }$

$\displaystyle \sf{ \implies \frac{dy}{dx} = 1 - \sqrt{a} \times {x}^{ - \frac{1}{2} } }$

$\displaystyle \sf{ \implies \frac{dy}{dx} = 1 - \sqrt{ \frac{a}{x} } }$

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