anonymous
  • anonymous
Find y' sinsqrt y -3x = 2 assuming that the equation determines a differentiable function f such that y=f(x)
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
you need to include brackets
anonymous
  • anonymous
\[1/2 (\sin y)^{-1/2}(\cos y)y'-3=0\] that is what i got but i dont know how to find y' now i.e. simplify
anonymous
  • anonymous
its not clear what the actual question is

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anonymous
  • anonymous
\[\sin \sqrt{y} - 3x=0\]
anonymous
  • anonymous
yeh
anonymous
  • anonymous
so sin( y^(1/2) ) -3x =2
anonymous
  • anonymous
(1/2)y^(-1/2) dy/dx cos(y^(1/2) ) -3 =0
anonymous
  • anonymous
or (sin y)^1/2
anonymous
  • anonymous
\[\frac{1}{2\sqrt{y}} \frac{dy}{dx} \cos(\sqrt{y}) -3 =0 \]
anonymous
  • anonymous
why'd we get rid of the sin? in the first term
anonymous
  • anonymous
solve that
anonymous
  • anonymous
why not ? lol
anonymous
  • anonymous
derivative of sin(f (x) ) = f'(x) cos(f(x) ) yes?
anonymous
  • anonymous
yes
anonymous
  • anonymous
differentiate the inside, multiply and channge the sin to cos
anonymous
  • anonymous
change
anonymous
  • anonymous
ok thank you i understand that was my fall down
anonymous
  • anonymous
you were thinking of \[\frac{1}{2} \sin^2 (\sqrt{y} ) -3x =2 \] I think
anonymous
  • anonymous
or something similiar

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