As this question contain some complicated degree, we can use logarithmic differentiation to slove this problem.
First, we take #ln# on both sides of the equation.
#y=(sinx)^lnx*(lnx)^cosx#
#ln(y)=ln[(sinx)^lnx*(lnx)^cosx]#
#lny=ln[(sinx)^lnx]+ln[(lnx)^cosx]#
#lny=lnx*ln(sinx)+ cosx*ln(lnx)#
Now, the problem of complicated degree are solved and we can differentiate both sides.
#d/dx(lny)=d/dx[lnx*ln(sinx)+ cosx*ln(lnx)]#
#1/y*dy/dx=d/dx[lnx*ln(sinx)]+ d/dx[cosx*ln(lnx)]#
#1/y*dy/dx=(lnx)*d/dx[ln(sinx)]+ln(sinx)*d/dx(lnx)+ cosx*d/dx[ln(lnx)]+ln(lnx)*d/dx(cosx)#
#1/y*dy/dx=(lnx)*1/sinx*cosx+ln(sinx)*(1/x)+ cosx*1/(lnx)*d/dx(lnx)+ln(lnx)*(-sinx)#
#1/y*dy/dx=(lnxcosx)/(sinx)+(ln(sinx))/x+ cosx*1/(lnx)*1/x-sinx ln(lnx)#
#1/y*dy/dx=(lnxcosx)/(sinx)+(ln(sinx))/x+ cosx/(xlnx)-sinx ln(lnx)#
#dy/dx=[(lnxcosx)/(sinx)+(ln(sinx))/x+ cosx/(xlnx)-sinx ln(lnx)]*y#
#dy/dx=[(lnxcosx)/(sinx)+(ln(sinx))/x+ cosx/(xlnx)-sinx ln(lnx)]*[(sinx)^lnx*(lnx)^cosx]#
Here is the answer. Hope this can help you :)
If you need further explaination, feel free to ask me.
