Here, #g(x) = lnx + sinx# and #h(x) = xlnx - x#. We must know the derivatives of both these functions to find #f'(x)#.
Starting with #g(x)#. The derivative of #lnx # is #1/x#, and the derivative of #sinx# is #cosx#.
By the sum rule, we have #g'(x) = 1/x + cosx#
Now for #h(x)#:
As you might have noticed, we will need to apply the product rule to #xlnx# in order to differentiate.
The derivative of #x# is 1 and the derivative of #lnx# is #1/x#.
Hence,# (xlnx)' = 1 xx lnx + x(1/x) = lnx + 1#
The derivative of #-x# is #-1#. Thus, #h'(x) = lnx + 1 - 1 = lnx#.
We can now apply the product rule to complex function #f(x)#.
#f'(x) = (1/x + cosx)(xlnx - x) + (lnx + sinx)lnx#
#f'(x) = (cancel(x)lnx)/cancel(x) + xlnxcosx - cancel(x)/cancel(x) - cos^2x) + ln^2x +lnxsinx#
#f'(x) = lnx + xlnxcosx - 1 + ln^2x + lnxsinx#
Hopefully this helps!
