can you help me get the y' of this equation?
x^1/2 + y^1/2 = a^1/2

Question

anonymous · Answer

By y' do you mean dy/dx?

anonymous · Answer

yep. dy/dx

anonymous · Answer

or inverse

anonymous · Answer

I would use implicit differentiation.  I get y'=(1/2)(x^(-1/2)+1/(a^(1/2)-x^(1/2))) although it may be wrong.

anonymous · Answer

Wait that's wrong, hold on...

anonymous · Answer

the book says the  answer is $$(-y ^1/2) / (x^1/2)$$

anonymous · Answer

but i don't know how the answer arrived to be like that.

anonymous · Answer

$$\sqrt{x}+\sqrt{y}=\sqrt{a}$$
$$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}y^'=0$$
$$\frac{1}{2\sqrt{y}}y^'=-\frac{1}{2\sqrt{x}}$$
$$y'=-\frac{\sqrt{y}}{\sqrt{x}}$$

anonymous · Answer

Yep that's right.  I got a minus sign wrong and then resubstituted to get rid of the y.  First diff implicitly to get 1/(2x^(1/2))+(y')/(2^(1/2))=0 etc.  What the person above said, lol

anonymous · Answer

why did the variable a turned to be equal to zero?

anonymous · Answer

in this case a is assumed to be a constant, whose derivative is 0

anonymous · Answer

because a^(1/2) is independent of x

anonymous · Answer

i see. thanks for the help. :) 
i'll give you two a medal.