Question

Find y' sinsqrt y -3x = 2

assuming that the equation determines a differentiable function f such that y=f(x)

Answer 1

you need to include brackets

Answer 2

\[1/2 (\sin y)^{-1/2}(\cos y)y'-3=0\] that is what i got but i dont know how to find y' now i.e. simplify

Answer 3

its not clear what the actual question is

Answer 4

\[\sin \sqrt{y} - 3x=0\]

Answer 5

yeh

Answer 6

so sin( y^(1/2) ) -3x =2

Answer 7

(1/2)y^(-1/2) dy/dx cos(y^(1/2) ) -3 =0

Answer 8

or (sin y)^1/2

Answer 9

\[\frac{1}{2\sqrt{y}} \frac{dy}{dx} \cos(\sqrt{y}) -3 =0 \]

Answer 10

why'd we get rid of the sin? in the first term

Answer 11

solve that

Answer 12

why not ? lol

Answer 13

derivative of sin(f (x) ) = f'(x) cos(f(x) ) yes?

Answer 14

yes

Answer 15

differentiate the inside, multiply and channge the sin to cos

Answer 16

change

Answer 17

ok thank you i understand that was my fall down

Answer 18

you were thinking of \[\frac{1}{2} \sin^2 (\sqrt{y} ) -3x =2 \]

I think

Answer 19

or something similiar