I can't speak for your calculator but the slope of the tangent line is dy/dx no matter what coordinate system you use.
Exactly what are you doing and what is the result?
Ok, I know the basic steps of finding the slope of polar curves. You just put in the equation and select "calculate" and then you have 3 options - value, dy/dx, and dr/dtheta. But I'm really confused about something. I thought dy/dx was the standard slope/derivation expression for polar curves because I've been doing polar/slope problems using dy/dx. What's up? Also, just how exactly do you find the slope of polar curves? I've been trying to use the calculator to find polar slopes but my answers NEVER match the book and they're not even remotely close most of the times.
Ok, I'll give you an example.
Q: Find the slope of 4cos(3theta) at theta = 2
Here's what I did. I make sure the calculator is in the degree and polar modes and I just input the equation in and graph it. After that, I click "2nd" then "calc" and then "dy/dx" and then I chose the value of "2" for the slope. My calculator gave me -2.823799 but the book says the answer is -0.2968.
By the way, I'm using the T1-83 calculator.
See any red flag or what?
Then go with radians. Except for elementary problems where you actually have triangles and angles are specified as being in degrees, trig functions are in terms of radians. For one thing, (sin(x))'= cos(x) and (cos(x))'= -sin(x) are only true if x is in terms of radians.
There are other ways of defining sine and cosine, by the way, in which the arguments are not angles at all! In those cases the x values correspond to radians.