Question

@zepdrix p= sec q + csc q/ csc q Find the derivative.

Answer 1

Is the denominator under all? or just the second term?

Answer 2

Under all!

Answer 3

Oh I guess that would make sense :3

Answer 4

I just figured you would cancel out the csc q but that was wrong. The correct answer is dp/dq= sec^2 q but I don't know how :/

Answer 5

\[\large p=\frac{\sec q+\csc q}{\csc q}\]Are we allowed to simplify before taking the derivative?

Answer 6

I think so, just as long as you reach the correct answer

Answer 7

Yah we can do some cancellation stuff the way you wanted to, let's split up the fractions first.\[\large p=\frac{\sec q}{\csc q}+\frac{\csc q}{\csc q}\]

So that gives us,\[\large p=\frac{\sec q}{\csc q}+1\]

Answer 8

For that first term, we'll want to rewrite each trig function in terms of their respective sines and consines.

Answer 9

\[\large \sec x=\frac{1}{\cos x}\]Right?
Remember cscx?

Answer 10

Yes!

Answer 11

\[\large p=\frac{\left(\dfrac{1}{\cos q}\right)}{\left(\dfrac{1}{\sin q}\right)}+1\]
So we get something like this, right bubs?

Answer 12

lol right! :)

Answer 13

Which simplifies to,\[\large p=\frac{\sin q}{\cos q}+1\]Which further simplifies to,\[\large p=\tan q+1\]

Answer 14

Is tan x a derivative you've gone over yet?
If it's not one you have memorized at this point, then you might be expected to take the derivative from the sine/cosine point.

Answer 15

Yes we have! So that would be sec^2 q!

Answer 16

Yay good job!

Answer 17

I know you want a break but are you up for two more? I promise those will be it for tonight! :x

Answer 18

Just post some questions, I'm sure someone will get to them!! :D
You can throw a @zepdrix in there if you want, and I'll come look later.

Answer 19

Or you can go to the lobby and look for the guys with big #'s by their name and @ those smarty pants. they can usually help XD

Answer 20

LOl okay! Thank you so much for your help! :) You're the best!

Answer 21

np \c:/