Verified answer

➥dy/dθ = dy/dr / dθ/dr

r-1 = sinθ

Take the derivative implicitly to find dθ/dr:

1 = cosθ dθ/dr

dθ/dr = secθ

Now start the conversion, but don't finish it...

r = 1 +sinθ

r = 1 + y/r

r² = r + y

y = r² - r

dy/dr = 2r-1

➥Therefore, dy/dθ = (2r-1) / secθ

dy/dθ = [2(1 + sinθ) - 1] /secθ

= [1 + 2sinθ] / secθ

or

cos θ + sin(2θ)

You can't.

You can't get dy/dθ to r(θ) because r(θ) doesn't have a y.

If you convert to Cartesian you don't have a θ.

If you want the first derivative of 1+sin(θ) that's dead easy.

1 is a constant so it just disappears so you're really just looking for the derivative of sin(θ).

The derivative of sin(θ) = cos(θ).

So r'(θ) = cos(θ)

or in the terminology you're using:

d(1+sin(θ))/dθ =cos(θ)

or

dy/dθ of y = 1+sin(θ) = cos(θ)

However you can't get dy/dθ if y is not defined.

Hope this helps.

y = (r)sin(θ)

r = 1 + sin(θ)

y = {1 + sin(θ)}sin(θ)

dy/dθ = cos(θ)sin(θ) + cos(θ)sin(θ) + cos(θ)

dy/dθ = sin(2θ) + cos(θ)

dy/dtheta = r '(theta) = cos(theta)