We are going to have to use more than one rule here, first the product rule. If we were to look at just "#x*sin(x)#", you might know how to differentiate the trigonometric function #sin(x)# and the linear function #x#. The product rule states:
#(f*g)'(x) = f'(x)*g(x)+f(x)*g'(x)#
so #(x*sin(x))' = 1*sin(x)+x*cos(x)#
Similarly #(x*sin(2/x))' = 1*sin(2/x)+x*(sin(2/x))'#
Looking at #(sin(2/x))'#, this require the use of the chain rule:
#dy/dx = dy/(du)*(du)/dx#
Let's say #u = 2/x# , so #sin(2/x) = sin(u)#
This means: #d/(du)sin(u) = cos(u)# and #d/dx(2*x^-1)=-2x^-2#
then we would get:
#(sin(2/x))' = (sin(u))' * (2/x)' = cos(u) * (-2x^-2) = -(2cos(2/x))/x^2#
Putting all this together, we get:
#(x*sin(2/x))' = sin(2/x)+x*(-(2cos(2/x))/x^2) = sin(2/x)-(2cos(2/x))/x#
PS. if you know how to differentiate and integrate #sin(x)# and #cos(x)#, then you know more than me, because I can't remember those to safe my life :)
