Given that, #f(x-y)=f(x)/f(y).......(star).#

#x=y, &, (star) rArr f(0)=f(y)/f(y)=1.......(1).#

#x=0, y=y, & (star) rArr f(-y)=f(0)/f(y)=1/f(y).........(2).#

#x=x, y=-y, & (star) rArr f(x+y)=f(x)/f(-y)=f(x)f(y)#

#:.," by "(2), f(x+y)=f(x)f(y).............(3).#

Knowing that, #f'(x)=lim_(h to 0) {f(x+h)-f(x)}/h,# we have, by #(3),#

#f'(x)=lim_(h to 0){f(x)f(h)-f(x)}/h=f(x){lim_(h to 0) (f(h)-1)/h}...(ast).#

Now, #f'(0)=p, (ast), & (1) rArr p=lim_(h to 0) (f(h)-1)/h...(4).#

Similarly, #f'(5)=q, (ast), & (4) rArr q=p*f(5)......(5).#

#"Finally, by "(ast), &, (4), f'(-5)=p*f(-5)#

#=p*(1/f(5)),............[because,(2)]#

#=p/(q/p),................[because, (5)]#

#:. f'(-5)=p^2/q,# as **Respected Cesareo R.,** has derived!

**Enjoy Maths.!**