#f(x-y)=f(x)/f(y)# and #f^'(0)=p,f^'(a)=q# then what is #f^'(-a)?#

We have:

# f(x-y) =f(x)/f(y) # and #f^'(0)=p,f^'(a)=q#, and seek #f^'(-a)#

Where (it is assumed) that #f# is a function of one variable. By direct substitution we have:

# x=y => f(x-x) = f(x)/f(x) #

# \ \ \ \ \ \ \ \ \ => f(0)=1# ..... [A]

# x=0 => f(0-y) = f(y)/f(y) #

# \ \ \ \ \ \ \ \ \ => f(-y)=1/f(y)# (using [A]) ..... [B]

# y=-y => f(x-(-y)) = f(x)/f(-y) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ => f(x+y)=f(x)/(1/f(y))# (using [B])

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ => f(x+y)=f(x) f(y)# ..... [C]

By the limit definition of the derivative, we have

# f'(x) = lim_(h rarr 0) \ (f(x+h)-f(x))/h #

# \ \ \ \ \ \ \ \ \ = lim_(h rarr 0) \ (f(x)f(h)-f(x))/h # (using [C])

# \ \ \ \ \ \ \ \ \ = lim_(h rarr 0) \ f(x)(f(h)-1)/h #

# \ \ \ \ \ \ \ \ \ = f(x) \ lim_(h rarr 0) \ (f(h)-1)/h #

We are given that #f'(0)=p#, so with #x=0#, we have:

# f'(0) = f(0) \ lim_(h rarr 0) \ (f(h)-1)/h #

# :. p = lim_(h rarr 0) \ (f(h)-1)/h # (using [A])

So, we can write:

# f'(x) = p \ f(x)# ..... [D]

We are also given that #f'(a)=q# and so with #x=a#, we have (using [D] that::

# q = p \ f(a) => f(a) = q/p#

And, from [B], we know that

# f(-y)=1/f(y) => f(-a) = 1/f(a) #

And again using [D] with #x=-a#, we have:

# f'(-a) = p \ f(-a)#

# \ \ \ \ \ \ \ \ \ \ \ \ \ = p /f(a)#

# \ \ \ \ \ \ \ \ \ \ \ \ \ = p / (q/p)#

# \ \ \ \ \ \ \ \ \ \ \ \ \ = p^2/q#