Differentiable function such that $f(x+y),f(x)f(y),f(x-y)$ are an arithmetic progression for all $x,y$ If $f$ is a differentiable function on $\mathbb{R}$ such that $f(x+y),f(x)f(y),f(x-y)$(taken in that order) are in arithmetic progression for all $x,y\in \mathbb{R}$ and $f(0)\neq0,$then
$(A)f'(0)=-1\hspace{1cm}(B)f'(0)=1\hspace{1cm}(C)f'(1)-f'(-1)=0\hspace{1cm}(D)f'(1)+f'(-1)=0$
My Try:
As $f(x+y),f(x)f(y),f(x-y)$ are in A.P.
$2f(x)f(y)=f(x+y)+f(x-y)$.........(1)
Put $x=0,y=0$ in the above equation
$2f^2(0)=2(0)$
$f(0)=1$ because $f(0)\neq0$
Differentiating the equation $(1)$,
$2f(x)f'(y)\frac{dy}{dx}+2f(y)f'(x)=f'(x+y)(1+\frac{dy}{dx})+f'(x-y)(1-\frac{dy}{dx})$
I am stuck here .Now i dont know how to go further.Please help me.
 A: Put $y=0$ to obtain $2f(0)f(x)=2f(x)$ and $x=0$ to get $2f(0)f(y)=f(y)+f(-y)$. Thus we get $2f(x)=2f(0)f(x)=f(x)+f(-x)\implies f(x)=f(-x)$.
Since therefore $f$ is even, it has to be answer (D), because:
$$
f'(-1)=\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to1}\frac{f(-x)-f(-1)}{-x+1}=-\lim_{x\to1}\frac{f(x)-f(1)}{x-1}=-f'(1)
$$
A: For $x=y$, you have $$f(x-x)+f(x+x)=2f(x)f(x)$$ or $$f(0)+f(2x)=2f^2(x)$$ or, $$2f^2(x)-f(2x)=1$$
Differentiating both sides with respect to $x$, we get
$$4f(x)f'(x)-2f'(2x) = 0 $$
Putting $x=0$, you have $$f'(0)=0$$
So options (a) and (b) are wrong.
As for (c) and (d), they are trying to check if $f'(x)$ is an even function or odd.
Use the fact that $$f(2x)=2f^2(x)-1$$ and $$f(-2x)=2f^2(-x)-1$$
Now assuming the function is even, i.e.$$f(2x)=f(-2x) \Rightarrow f^2(x)=f^2(-x) \Rightarrow f(x)=f(-x)$$ That is, the equation supports that the function is even.
But assuming the function is odd, $$f(2x)=-f(-2x)$$ $$\Rightarrow 2f^2(x)-1=1-2f^2(-x)$$ $$\Rightarrow 2[f^2(x)+f^2(-x)]=2$$ $$\Rightarrow [f^2(x)+f^2(-x)]=1 \not \Rightarrow f(x)=-f(-x)$$ That is, the equation does not support that the function is odd.
So your answer is (d).
A: 
$f(x+y) ,f(x)f(y),f(x-y)$ are an arithmetic progression 

All nontrivial twice-differentiable solutions are $f(x) = \frac{e^{Ax} + e^{-Ax}}{2}$ for $A$ real or pure imaginary.
The functional equation is $f(x+y) + f(x-y) = 2f(x)f(y)$.
For small $y$, twice-differentiable solutions satisfy $$2f(x) + y^2 f''(x) + o(y^2) = 2f(x)f(y).$$  If there is any point $x$ where $f''(x)f(x) \neq 0$, that implies $2f(x)(f(y)-1) = O(y^2)$ near $y=0$.  It is then possible to divide by $y^2$ and the limit as $y \to 0$ necessarily exists:
$$ f''(x) = 2f(x) \lim_{y \to 0} \frac{f(y)-1}{y^2} = f(x) f''(0)$$
so that $f''(x) = Cf(x)$ for constant $C$ (and $f(0)=1$ and $f'(0)=0$).
Taking $A^2 = C$ yields the stated solutions.
A: $$2f(x)f(y)=f(x+y)+f(x-y)$$
Clearly $$f(x)=\cos x$$ satisfies above equation.
