So we have
#y = cos(x)*e^(-x)*sqrt(3x)#
To simplify things, let's say #e^(-x)sqrt(3x) = u#
So we have
#y = u*cos(x)#
Using the product rule we have
#dy/dx = ud/dx(cos(x)) + cos(x)(du)/dx#
#dy/dx = -u*sin(x) + cos(x)(du)/dx#
And now we differentiate #u#
#u = e^(-x)* sqrt(3x) = e^(-x)*(sqrt(3)x^(1/2))#
#(du)/dx = e^(-x)d/dx(sqrt(3)x^(1/2)) + sqrt(3x)d/dx(e^(-x))#
#(du)/dx = e^(-x)*sqrt(3)/2x^(-1/2) -sqrt(3x)*e^(-x)#
#(du)/dx = e^(-x)((sqrt(3))/(2sqrt(x)) - sqrt(3x))#
Using some algebra to change that we have
#(du)/dx = e^(-x)((sqrt(3x))/(2x)-sqrt(3x))#
#(du)/dx = e^(-x)*sqrt(3x)(1/(2x) - 1)#
#(du)/dx = (u(1-2x))/(2x)#
Substitute that back in the original derivative
#dy/dx = -u*sin(x) + cos(x)*u\*(1-2x)/(2x)#
#dy/dx = u(cos(x)*(1-2x)/(2x) - sin(x))#
#dy/dx = e^(-x)\*sqrt(3x)(cos(x)*(1-2x)/(2x)-sin(x))#