Differentiating the #4sinx# is significantly easier than the #4x^x#.

If #f(x)=4sinx+4x^x#, #f'(x)=d/dx(4sinx)+d/dx(4x^x)#

#f'(x)=4d/dx(sinx)+4d/dx(x^x)#

#d/dx(sinx)=cosx# (important one to learn!)

#d/dx(x^x)# can not be computed using the generalised power rule for differentiation, such that: #d/dx(x^x)!=x(x^(x-1))# since x is not a constant.

We can compute it however, by using the following rule:

#f^g=e^(glogf)#, where #f# denotes #f(x)#, #g# denotes #g(x)# and #log# is the natural logarithm.

By extension: #x^x=e^(xlogx)# and therefore:

#d/dx(x^x)=d/dx(e^(xlogx))#

The whole point of this is that it's possible to take the derivative of this function easily. This is because the derivative of #e^x# is just #e^x#, so we only need to multiply through by the derivative of what's in the power (as given by the chain rule).

Therefore #d/dx(e^(xlogx))=e^(xlogx)(d/dx(xlogx))#

But there's no point leaving the result like that since we can just go back to #x^x# from #e^(xlogx)#.

Also, #d/dx(xlogx)=1+logx#

So finally, #f'(x)=4cosx+4x^x(1+logx)#