anonymous
  • anonymous
Solve the equation explicitly for y and and differentiate to get y' in terms of x. (x^(1/2))+(y^(1/2))=1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\sqrt{x}+\sqrt{y}=1\]
anonymous
  • anonymous
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.
freckles
  • freckles
\[(1-\sqrt{x})^2 \neq 1+x\]

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anonymous
  • anonymous
oh it doesnt?
freckles
  • freckles
\[(1-\sqrt{x})^2=(1-\sqrt{x})(1-\sqrt{x})=1-\sqrt{x}-\sqrt{x}+x=1-2 \sqrt{x}+\sqrt{x}\] but I would just leave it as (1-sqrt(x))^2 and use chain rule
anonymous
  • anonymous
ok one second let me change that up and see what I get now.
freckles
  • freckles
oops I made a type-0
freckles
  • freckles
that last term was suppose to be x
anonymous
  • anonymous
yea the last sqrt x should be just x
freckles
  • freckles
\[1-2 \sqrt{x}+x\]
anonymous
  • anonymous
I got \[\frac{ 1-\sqrt{x} }{ \sqrt{x} }\]
anonymous
  • anonymous
is that what you got as well?
freckles
  • freckles
hmmm I got the opposite of that
freckles
  • freckles
which just means our answers are off by a constant multiple of 1
anonymous
  • anonymous
oh wait I forgot the negative in my answer
freckles
  • freckles
\[\sqrt{x}+\sqrt{y}=1 \\ \sqrt{y}=1-\sqrt{x} \\ y=(1-\sqrt{x})^2 \\ y'=2(1-\sqrt{x}) \cdot (0- \frac{1}{2 \sqrt{x}}) \\ y'=\frac{-2(1-\sqrt{x})}{2 \sqrt{x}} =\frac{-(1-\sqrt{x})}{\sqrt{x}}\]
anonymous
  • anonymous
yup now I got that as well
freckles
  • freckles
cool stuff
anonymous
  • anonymous
alright thank you again :D
freckles
  • freckles
np you are welcome again

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