Can someone explain? (1 Viewer)

[king_of_boredom]

what is the second derivative of y = tan x? (y'' = ?) and if you know the answer could you explain to for me? thanks.

 

[Trev]

y = tanx
y' = sec²x = 1/(cos²x)

Using y' = (vu' - uv')/v² [u=1; v=cos²x]
y'' = (cos²x.0 - 1.2.cosx.-sinx)/cos⁴x
= (2sinx)/(cos³x)

Or

y' = (cosx)^-2
y'' = -2(cosx)^-3.-sinx
= 2sinx(cosx)^-3

 

[king_of_boredom]

thanks dude, the actual question in the exam was

If y = tan x show that y'' - 2y^3 - 2y = 0

and i ended up getting it right! and i got 90% in that assessment! woohoo i have a huge lead on every1 in both 2 and 3 unit.....i guess all the hard work i put in paid off

 

[haboozin]

thanks dude, the actual question in the exam was

If y = tan x show that y'' - 2y^3 - 2y = 0

and i ended up getting it right! and i got 90% in that assessment! woohoo i have a huge lead on every1 in both 2 and 3 unit.....i guess all the hard work i put in paid off

A question like this
that asks prove something equals 0 and you get it right means u would know at the exam time that you have got it right if u have proved it...

so why did u ask how to do it, if u did it and u got the right answer...

 

[currysauce]

he can ask what he likes

 

[king_of_boredom]

A question like this
that asks prove something equals 0 and you get it right means u would know at the exam time that you have got it right if u have proved it...

so why did u ask how to do it, if u did it and u got the right answer...

no what actually happened is after i did the test, i wasn't sure if i got it right. so i came on and asked what the second derivative of y = tanx was. funny thing is, i got the test back and got full marks for it, even without finding the second derivative of y = tanx. what i did was find 2y^3, and then 2y, plug them both in and find a general eq'n for y''. then i just put them all together and turns out i got the right answer.