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There are a whole set of functions that differentiate to give themselves. But they are all multiples of each other.
Show this is equal to zero which means that h(x) = constant.
I think that should enable you to do what you want.
That was an awesome proof,but i need one more favor,if f'(x)=f(x) and f(0)=1 than f(x)=a^x for some constant a,how do i prove this?
I really like that proof. Really elegant.
I suspect there's a circular argument lurking here as power series probably depend on natural logs somewhere, but maybe it's ok.
This is how I do powers and logs:
at (0,1), the derivative is:
Even though I don't know what that is, it will have a value; let's say k.
Now the derivative at other points
So all graphs in the family have the property that the gradient function at x is a^x times the gradient at (1,0)
In the family there will be one value of a for which k = 1
Call that one a = e
Taking logs base e for the first expression:
Differentiating wrt x
which means we now know the value of k ... and
so [still working on this last bit but I think I'll post before I lose it all]
No good. I seem to be stuck here because if k - ln a this becomes ln x and I was trying hard to avoid that. I seem to have gone too far and proved the log is base e. I'll come back to it later after a think.
Proof from first principles will be better
Help me prove this:for some constant a[i cannot use the fact this is ln(x)]