Find d^48/d^48 (xsinx) by finding the first few derivatives and observing the pattern that occurs.?

Question

Hi, could anyone help me with the following math problem? Thanks!

Find d^48/d^48 (xsinx) by finding the first few derivatives and observing the pattern that occurs.

Answer 1

xsinx

d^1/dx^1 (xsinx) = sinx + xcosx

d^2/dx^2 = cosx + cosx - xsinx = 2cosx - xsinx

d^3/dx^3 = -2sinx - xcosx - sinx = -3sinx - xcosx

d^4/dx^4 = -3cosx + xsinx - cosx = -4cosx + xsinx

With those as a guide you can determine that the solution is...

********d^48/dx^48 (xsinx) = -48cosx + xsinx********

With sin/cos the derivative is a 4 part cycle...

d/dx (sinx) = cosx
d/dx(cosx) = -sinx
d/dx(-sinx) = -cosx
d/dx(-cosx) = sinx

We know it's d^48/dx^48, so the coefficient on the first term has to be +/-48. 48 falls on the 4th portion of the cycle, so therefore the first term is a -cosx. Combining those two we have that the first portion of the term is -48cosx. Similarly we can look at our d^4/dx^4 and see there is a positive xsinx tacked on, so that is also included in the final solution for the d^48/dx^48.

Answer 2

The pattern goes in cycles of four.

y = xsinx

d^(4n+1)/dx^(4n+1) = (4n+1)sinx + xcosx
d^(4n+2)/dx^(4n+2) = (4n+2)cosx - xsinx
d^(4n+3)/dx^(4n+3) = -(4n+3)sinx - xcosx
d^(4n)/dx^(4n) = -(4n)cosx + xsinx

The pattern then repeats.

48 is of the form 4n so

d^(48)/dx^(48) [xsinx] = -48cosx + xsinx