Recent content by tad.confused

I need to find the derivative of the square root of (2x+1) (not sure how to do square root symbol here, sorry) I understand that the square root of (2x+1)= (2x+1)^(1/2), but I am getting a little confused on how to continue from there.

I get it now! Thank you sooooo much!:biggrin:

Oh! wow didn't even think about that! thanks! Oh, so secx=1/cosx and its the same for sec2x=1/cos2x? also, thank you for the tip about the subscript, so much easier :smile:

also, if I did the math right, I went from (sec^2x+(2sec^2xtanx)+sec^2x) and then dividing it all by sec^2x and canceled all of the sec^2x's and got 2+2xtanx This is *kind of* close to what the back of my book says which is (2+2xtanx)/cos^2x=(2cosx+2xsinx)/((cos^3)x)

ok, so i get what was done, but the answer that is in the back of the book (and I understand that they are wrong sometimes) says that it is (2cosx+2xsinx)/((cos^3)x)

Oh, I had it right on the paper, but I typed the first derivative wrong. oops.

Homework Statement Find the second derivative (y") of y=xtanx. The attempt at a solution I got the first derivative (y') y=xtanx y'=x(secx)+tanx I started the second derivative and got stuck y"=xsec^2x+tanx