Trig derivative; stuck midway please help ! :)

Question

anonymous · Answer

$$\large y= \frac{ tanx-1 }{ secx }$$

loser66 · Answer

take derivative of a quotient, that's it

anonymous · Answer

i am but i have a mistake midway, can you please help me ?

anonymous · Answer

$$\large y'= \frac{ (secx)(\sec^2x)-(tanx-1)(secxtanx) }{ \sec^2x }$$

anonymous · Answer

$$\large y'=\frac{  \sec^3x- ?????? }{ \sec^2x }$$

loser66 · Answer

???? = tanx^2x secx -tan x (just multiply)

anonymous · Answer

Ohhhhh, how stupid of me -.-

anonymous · Answer

@Loser66  thanks !! :)

anonymous · Answer

$$\large y'= \frac{ \sec^3x-(secstan^2x-secxtanx) }{ \sec^2x }$$

loser66 · Answer

factor sec out, you can go ... somewhere else,hihi... try

anonymous · Answer

(secstan2x−secxtanx) <<<  is that negative of positive inside of the brackets?

loser66 · Answer

$$y'=\frac{secx(sec^2x-tan^2x+tanx)}{sec^2x}$$
$$=\frac{1+tanx}{secx}= cosx +sinx$$
hopefully no mistake

loser66 · Answer

hi friend, you can go this way, 
$$\frac{tanx-1}{sec}=\frac{tanx}{sec}-\frac{1}{sec}$$
$$=sinx -cosx$$
then take derivative that stuff to get 
$$y'=cosx+sinx$$
sorry for lagging, I just follow yours without thinking about this stuff