#f(x-y)=f(x)/f(y)# and #f^'(0)=p,f^'(a)=q# then what is #f^'(-a)?#
1 Answer
# f'(-a) = p^2/q#
Explanation:
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
# 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) = 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
# 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
# f'(-a) = p \ f(-a)#
# \ \ \ \ \ \ \ \ \ \ \ \ \ = p /f(a)#
# \ \ \ \ \ \ \ \ \ \ \ \ \ = p / (q/p)#
# \ \ \ \ \ \ \ \ \ \ \ \ \ = p^2/q#