fifth derivative

now what is that??

actually i have just now started this chapter ..,, it's been a couple of classes

how does cross multiplying help??
mam always gives some or the other kind of tricky questions ....i don't know what's the trick here

c
y(1+x^2) =1
so

difrentiating y keepin d othr constant
then difrntating d othr aprt keepin y constant

dy/dx (1+x^2) +2x y =0

diff again

do this 5 tyms :)

hehe dude rohanjiii dis fifth derivative and all doesnt come for any exams..........maximum which came is third derivative,........or often second derivative problems are asked for any competitive exams.........for ur information...........

i have a better way .....1+x²=x²-i² i =√-1 ,

(x+i)(x-i)

1/(x+i)(x-i) =y

now let y= A/(x+i) + B/ (x-i)
we can easily calculate A and B ....which come out to be A= 1/2i and B= -1/2i

so y=(1/2i) [1/(x-i) - 1/(x+i)]
if you now differntiate it ,, it will be far more easier ....:):):):):)

wow what a method ...nice

thanks