Trig Differentiation Question

$x = 4sin(2y+6)$
Find $\dfrac{dy}{dx}$ in terms of x

Now I can find it in terms of y it's just:
$8cos(2y+6)$
But idk how to get it in terms of x, do I substitute it in somewhere along the line?

That's only dx/dy
$\cos y = \sqrt{1-\sin^2y}$ for suitable $y$, so you can use it here.

Okay that's handy I think? But I don't know how to get the y out of the sine function? Is there a trig rule I'm not remembering?

This comes from $\sin^2 x + \cos^2 x \equiv 1$

You get $4\cos (2y+6) = \sqrt{16-16\sin^2(2y+6)}$ so $4\cos (2y+6) = \sqrt{16-x^2}$

Okay I think I've got what you're getting at, I think you had sine and cos round the wrong way in your post but you just swap them and you get:
$x=\sqrt{16-16cos^2(2y+6)}$
and I know that:
$\dfrac{dx}{dy} = 8cos(2y+6)$
So I manipulate the first equation until it looks like the second to get:
$\dfrac{dx}{dy} = \dfrac{\sqrt{16-x^2}}{2}$
Is this the right idea and have I messed up along the way at all

(Also thanks a bunch for pointing me in the right direction instead of just giving me the answer)

RDKGames;76216720]It doesn't matter which way they are since the equalities will still be true. In this case, you want to express cos in terms of x which is what I've had it written as.

You ended up with $\dfrac{dx}{dy} = 8 \cos (2y+6)$ which is just $2\sqrt{16-x^2}$, not half.

Okay so I started from the beginning again and got $\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{16-x^2}}$

Is this right (finally?)