1why isn't anybody interested in answering this question??????
if f(x)= sin-1[ (3x4+6x2-1)/(x2+1)3] find dy/dx at x2= tan (pi/8)
i tried this method
(3x4+6x2-1 )/(x2+1)3 = 3 sin a -4 sin3a=sin 3a where sin a= 1/(1+x2)
now x2=tan pi/8
so we have sin a =cos2pi/8....
and f(x) becomes sin-1(sin 3a)......
but we can write it as 3a or pi-3a(etc) only when we know the Value of cos2pi/8 which comes out to be 1/2 + 1/2√2 ....for which we do not have any perfect angle a .....now what should i write sin-1(sin3a) as .....if we directly differentiate it ,,,it becomes hard to do that ..now what should i do???
if you have some other method please post it...please!!! :):):)
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4 Answers
62i din understand how it became harder..
f(x)=a
f'(x)=da/dx
sin a= 1/(1+x2)
cos a da/dx = -2x/(1+x2)2
now it is very simple!
I did not understand where you are getting stuck!
