Question

Can someone help me find the derivative of the following please: y = √3x + 2 cos(x), 0 ≤ x < 2π

Answer 1

I know I can rewrite it and have \[y=3^{1/2}x+2\cos (x)\]

Answer 2

\[y'= \frac{ 3 }{ 2 }^{-1/2}x-2\sin x\]

Answer 3

I think I forgot a step but, I don't know for sure.

Answer 4

\[\frac{d}{dx}[\sqrt{3}\cdot x]=\sqrt{3}\]

Answer 5

okay so would I get \[y'=\sqrt{3}-2\sin x\]

Answer 6

yes

Answer 7

oh okay I was way over thinking that. Thanks a bunch. I have been working on calculus homework for the past 2 days. I think I need to take a break. I should have known that.