need help in C3

[T.Q]

Given that x = 4 sin (2y + 6), find dy/dx in terms of x.

my answer to the Q was  dy/dx= 1/8cos(2y+6)     
but i didnt know how to write it in term of x can someone help me plz

[XFlufferzX]

x = 4 sin (2y + 6)

Sin (2y +6) = x/4

where x = opp.
         4 = hyp.
To find adjacent uuse Pythagoras theorem

adj = sqrt 16 - x^2

Cos ( 2y +6) = adj / hyp
                  = (sqrt. 16 - x^2)/4

dy/dx = 1/8cos(2y+6)
Simplify

1/8 . 1/cos(2y+6)

1/8 . 1/ sqrt. 16 - x^2 / 4

= 1/2 sqrt 16 - x^2

eh If yuh couldnt understand wat I did let me know
check da attached paper

[T.Q]

thank you:)

[astarmathsandphysics]

See this for implicit differentiation

http://www.astarmathsandphysics.com/a_level_maths_notes/C3/a_level_maths_notes_c3_implicit_differentiation.html

I am just about to finish the C3 notes. Start C4 tomorrow

[T.Q]

See this for implicit differentiation

http://www.astarmathsandphysics.com/a_level_maths_notes/C3/a_level_maths_notes_c3_implicit_differentiation.html

I am just about to finish the C3 notes. Start C4 tomorrow

thank you so much ,
very helpful