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

1 Answer
Jun 22, 2018

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 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