1- e^x cosx ..... hariya babu -ve sign kaha gaya.... :)
11 Answers
1708as we know that e(ix) = cosx + i.sinx.
and e(-ix) = cosx - i.sinx.
so sinx = (eix + e-ix) / 2.
I think this may help you.....
1y=exsinx ( given )
dydx= exsinx + excosx
d2ydx2=exsinx + excosx + excosx - exsinx
d2ydx2=2(excosx)
d3ydx3=2(excosx - exsinx)
d4ydx4 =2(excosx - exsinx - exsinx - excosx)
d4y dx4= 2(-2exsinx)= -4exsinx
NOW TAKING -4 AS COMMON WE CAN SEE THAT WE HAVE TO DO AGAIN DERIVATIVE OF exsinx ,so AFTER FOURTH DERIVATIVE NEXT FOURTH DERIVATIVE WILL BE ,
d8y
dx8=-4d4xdx4
d8y
dx8=-4(-4exsinx)=16exsinx
d10ydx10=16(d2y/dx2)
d10ydx10=16(2excosx)
d10ydx10=32 excosx
is it correct now steps seems to be long but when u do roughly, trick works in seconds
11@nihal... hey how come..
d8y/dx8 =(d4y/dx4)2...
then d2y/dx2 should be = (dy/dx)2..
is it so???
1hahaha..... no its not so...:P
y = exsinx
=> dydx = exsinx + excosx = ex(sinx + cosx) = √2exsin (x + Π4)
=> d2yd2x = (√2)2ex sin (x + 2.Π4)
.
.
...proceeding this way,
dnydxn = (√2)nexsin (x + nΠ4)
so now put n=10 and get the ans 32 excosx :)
