I'm pretty sure you're looking for the derivative, seeing as this is posted under "Differentiating Trigonometric Functions".
Let #f(x) = xcos(sqrtx)#. This is a product of the functions
#x#, and #cos(sqrtx)#. So, using the product rule:
#(df)/dx = (x)' cos(sqrtx) + x[cos(sqrtx)]'#
#=cos(sqrtx) + x[cos(sqrtx)]'#.
Now we need to find the derivative of #cos(sqrtx)#:
The chain rule states that, if #y# is a function of #u#, and #u# is a function of #x#, then
#(dy)/(dx) = (dy)/(du) (du)/(dx)#.
Let #y = cos(sqrtx)# and #u = sqrtx#. Then #y = cosu#:
#(dy)/(dx) = -sinu * u' = -sinu * 1/(2sqrtx) = -(sin(sqrtx))/(2sqrtx)#
Therefore,
#(df)/(dx) = cos(sqrtx) - (xsin(sqrtx))/(2sqrtx)#