Question

f(x)= 4cscx-cotx

Answer 1

Are you looking for the derivative or just solving the equation?

Answer 2

looking for the derivative!

Answer 3

Well, I have never done derivatives of csc or cot before, just sin and cos. But what I can say is that tan is just sin/cos, and cotan is the reverse so that it is cos/sin. So perhaps you should use quotient rule on that part of the problem? I'm not sure if that's right, just trying to be helpful until one of the Helpers comes along to help you out. :)

Answer 4

Oh, and the derivative of sin is cos, and the derivative of cos is -sin.

Answer 5

thank you so much!

Answer 6

No problem! I don't think that I helped too much, but I tried haha. I also remembered today that csc is equal to 1/sin. So basically you are looking at:\[4\frac{ 1 }{ \sin }-\frac{ \cos }{ \sin }\]
Which, with simple multiplication and combining them seeing as the denominators are the same, are:
\[\frac{ 4-\cos }{ \sin }\]
And from there, it's just the Quotient rule! \[\frac{ ba'-ab' }{ b ^{2} }\]

Answer 7

If you have any more questions, let me know. :)