Question

Solve the equation explicitly for y and and differentiate to get y' in terms of x.
(x^(1/2))+(y^(1/2))=1

Answer 1

\[\sqrt{x}+\sqrt{y}=1\]

Answer 2

So this is what I thought they are trying to get at:
\[\sqrt{y}=1-\sqrt{x}\]
square root both sides to take other the square root on y.
\[[\sqrt{y}=1-\sqrt{x}]^{2}\]
\[y=1+x\]
\[y'=1\]
This is wrong.

Answer 3

\[(1-\sqrt{x})^2 \neq 1+x\]

Answer 4

oh it doesnt?

Answer 5

\[(1-\sqrt{x})^2=(1-\sqrt{x})(1-\sqrt{x})=1-\sqrt{x}-\sqrt{x}+x=1-2 \sqrt{x}+\sqrt{x}\]
but I would just leave it as (1-sqrt(x))^2
and use chain rule

Answer 6

ok one second let me change that up and see what I get now.

Answer 7

oops I made a type-0

Answer 8

that last term was suppose to be x

Answer 9

yea the last sqrt x should be just x

Answer 10

\[1-2 \sqrt{x}+x\]

Answer 11

I got \[\frac{ 1-\sqrt{x} }{ \sqrt{x} }\]

Answer 12

is that what you got as well?

Answer 13

hmmm I got the opposite of that

Answer 14

which just means our answers are off by a constant multiple of 1

Answer 15

oh wait I forgot the negative in my answer

Answer 16

\[\sqrt{x}+\sqrt{y}=1 \\ \sqrt{y}=1-\sqrt{x} \\ y=(1-\sqrt{x})^2 \\ y'=2(1-\sqrt{x}) \cdot (0- \frac{1}{2 \sqrt{x}}) \\ y'=\frac{-2(1-\sqrt{x})}{2 \sqrt{x}} =\frac{-(1-\sqrt{x})}{\sqrt{x}}\]

Answer 17

yup now I got that as well

Answer 18

cool stuff

Answer 19

alright thank you again :D

Answer 20

np you are welcome again