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) andf^'(0)=p,f^'(a)=q , and seekf^'(-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