Differentiate

ma ans. is negetive of harry's answer

- e^x cosx ..... hariya babu -ve sign kaha gaya.... :)

as we know that e^(ix) = cosx + i.sinx.
and e^(-ix) = cosx - i.sinx.

so sinx = (e^ix + e^-ix) / 2.

I think this may help you.....

y=e^xsinx ( given )

dydx= e^xsinx + e^xcosx

d²ydx²=e^xsinx + e^xcosx + e^xcosx - e^xsinx

d²ydx²=2(e^xcosx)

d³ydx³=2(e^xcosx - e^xsinx)

d⁴ydx⁴ =2(e^xcosx - e^xsinx - e^xsinx - e^xcosx)

d⁴y dx⁴= 2(-2e^xsinx)= -4e^xsinx

NOW TAKING -4 AS COMMON WE CAN SEE THAT WE HAVE TO DO AGAIN DERIVATIVE OF e^xsinx ,so AFTER FOURTH DERIVATIVE NEXT FOURTH DERIVATIVE WILL BE ,

d⁸y
dx⁸=-4d⁴xdx⁴

d⁸y
dx⁸=-4(-4e^xsinx)=16e^xsinx

d¹⁰ydx¹⁰=16(d²y/dx²)

d¹⁰ydx¹⁰=16(2e^xcosx)

d¹⁰ydx¹⁰=32 e^xcosx

is it correct now steps seems to be long but when u do roughly, trick works in seconds

@nihal... hey how come..

d⁸y/dx⁸ =(d⁴y/dx⁴)²...

then d²y/dx² should be = (dy/dx)²..

is it so???

hahaha..... no its not so...:P

y = e^xsinx

=> dydx = e^xsinx + e^xcosx = e^x(sinx + cosx) = √2e^xsin (x + Π4)

=> d²yd²x = (√2)²e^x sin (x + 2.Î 4)
.
.
...proceeding this way,

dⁿydxⁿ = (√2)ⁿe^xsin (x + nÎ 4)

so now put n=10 and get the ans 32 e^xcosx :)

thats a better method than the previous one...