Question:
A function f:R→R satisfies the equation f(x y)=f(x)f(y) for all x,yϵR , f(x)≠0 suppose that the function is differentiable at x=0 and f'(0)=2 prove that f'(x)=2 f(x)
Answer:
Given, f(x + y) = f(x) * f(y)
This property can be satisfied by a function of type k * ax
If f(x) = ax
Then,
f(x + y) = ax+y
=> f(x + y) = ax * ay
=> f(x + y) = f(x) * f(y)
Now, f(x) = ax
=> df(x)/dx = ax * log a
Since, df(x)/dx = 2
=> 2 = a0 * log a
=> log a = 2
Hence, df(x)/dx = ax * log a
=> df(x)/dx = ax * 2
=> df(x)/dx = 2f(x)
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