I came across this problem :

y = 7x (cosx)^x/2 ;

where I'm to find y'. So with the understading that (cosx) is raised to a variable power of x/2, I get an idea of applying Logarithmic Differentiation as follows :

Iny = In [7x (cosx)^x/2],

then apply the Log Law saying that the Log of a Product is the Log of the Sum to have :

Iny = In (7x) + x/2 In(cosx), then go on to finding dy/dx by finding the derivative of the sum,in which I apply the Product Rule to find dy/dx of x/2 In(cosx) to get :

dy/dx Iny = dy/dx[In 7x] + [In(cosx)dy/dx (x/2) + (x/2) dy/dx In(cosx)]

=> dy/dx Iny=(1/7x)7 + [In(cosx)/2 - x(sinx)/2(cosx)]

=> dy/dx Iny=(1/x) + 1/2[In(cosx) - x(tanx)]

=> dy/dx = y {(1/x) + 1/2[In(cosx) - x(tanx)]},

Then by substituting the given y value, I obtain :

dy/dx = [7x(cosx)^x/2]{(1/x) + 1/2[In(cosx) - x(tanx)]}:];

Which I regard as my solution. Do I need to simplify any further ?