Maths differentiation problem?

Question

x=3sin y
dx/dy in terms of y is: 3 cos y
hence show dy/dx= 1/(9-x^2)^1/2
and also find a function of x with this derivative?

im not sure how the answer of dy/dx is found though eliminating the trig expressions, logical steps with explanation would be great. thanks :)

could you tell me exactly how it goes from (1-sin^2)^1/2 to (9-x^2)^1/2. this is the part im not getting thanks

Answer 1

dy/dx
= 1 / (dx/dy)
= 1 / (3cos y)
= 1 / [ 3 sqrt ( 1 - sin^2(y))]
= 1 / [ sqrt ( 9 - (3sin y)^2) ]
= 1 / ( 9 - x^2 )^( 1/2 )

dy = [ 1 / ( 9 - x^2 )^( 1/2 ) ] dx
Integrating,
y = arc sin ( x/3 )

Answer 2

dy/dx = 1/3cosy
Since siny =x/3, cosy = (1 -sin^2y)^1/2 or (1-x^2/9)^1/2 =
1/3*( 9 - x^2)^1/2
So dy/dx =1/3[1/3(9 - x^2)^1/2]
Cancel the 3 and 1/3 in the denominator and you have your answer.
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Answer 3

x=3sjny
dx/dy=3cosy
cos^2y=1-sin^y
=1-x^2/9
=(9-x^2)/9
cosy=sqrt(9-x^2)/3
dx/dy=sqrt(9-x^2). ANS.

Answer 4

dy/dx=1/(dx/dy) (Inverse function theorem)=1/3cosy
as sin^2 y +cos^2y = 1 cos y =+-sqrt(1-sin^2y) =+-(1-x^2/9)^1/2
so dy/dx = +-1/(9-x^2)^1/2 ( you can´t avoid the+-) unless you limit the range of sin y
dx/(9-x^2)=+-dy so y=+-arcsin(x/3)+C