The derivative of function under constraint

[Raghav Gupta]

I can also prove the inconsistency by taking,
f(x) = (f(5))^x, it is given in question that f(5) = 2.
Now I am going to show that if f(5) = 2 then f'(0) ≠ 3
Differentiating,
$f'(x) = f(5)^x ln f(5)$
Putting x = 0 and we know from question f(5) = 2
We see f'(0) = ln2 ≠ 3
So the values are dependent.
Is this also a way of correctly thinking?

 

[Ray Vickson]

Why f(x) is only equal to f((1))^x?
As Sammy is saying it could be any constant.
So we can also write f(x) =(f(2))^x ?

It's looking like I'm going round and round if you see my encounter with Dick in this thread.
So if we take
$h\to 0$
Then f'(x)= f(x) \frac{f(h) - f(0)}{h}
Now what to do?

No: this equation is false. For finite $h > 0$ the right-hand-side is NOT equal to $f'(x)$. Derivatives involve a limit, not just a difference-ratio.

 

[Raghav Gupta]

No: this equation is false. For finite $h > 0$ the right-hand-side is NOT equal to $f'(x)$. Derivatives involve a limit, not just a difference-ratio.

But if you see I have written $h\to 0$

 

[Ray Vickson]

But if you see I have written $h\to 0$

No, you have not written it properly and in the correct place. You should have written
f'(x) = f(x) \, \lim_{h \to 0} \frac{f(h) -f(0)}{h}
THAT would be a correct statement.

Do you see how it is different from what you wrote? (This not mere quibbling; if you are not careful to write things correctly, they can come back later to cause you to make difficult-to-trace errors. I am trying to help you develop good habits that will be useful to you later, when problems are a lot more complicated.)