# Homework Help: The derivative of function under constraint

1. May 12, 2015

### Raghav Gupta

1. The problem statement, all variables and given/known data

If f(x+y) = f(x)f(y) for all x and y and f(5) = 2, f'(0)=3, then f'(5) is
a) 5
b) 6
C) 0
d) None of these
2. Relevant equations

Don't know which equation to apply. I was thinking of Rolle's Theorem and mean value theorem here but it is not helping.
3. The attempt at a solution
Found f(0) = 1

2. May 12, 2015

### pasmith

You're not actually told that the function is differentiable anywhere other than at zero, so the first thing to do is prove that $f'(5)$ actually exists. Therefore you should start from $$f'(5) = \lim_{h \to 0} \frac{f(5 + h) - f(5)}{h}.$$

3. May 12, 2015

### HallsofIvy

Notice also that, because f(x+ y)= f(x)f(y), f(x+ 0)= f(x)= f(x)f(0) so we must have f(0)= 1. That helps!

4. May 12, 2015

Is the answer 6?

5. May 12, 2015

### SammyS

Staff Emeritus
I think we're waiting for OP to respond.

6. May 13, 2015

### Raghav Gupta

But I have already shown that in my attempt of post 1. How it is helping me?
Now what should I do?
I am getting a 0/0 form on right side and L Hopital rule also is not beneficial here.

7. May 13, 2015

### Dick

Derivative limits always have a 0/0 form. Use f(x+y)=f(x)f(y).

8. May 13, 2015

### Raghav Gupta

So, I am getting,
$$f'(5) = \lim_{h \to 0} \frac{f(5)f( h) - f(5)}{h}.$$
Now f(5) = 2 and f(0)=1, so f(h) = 1
Again getting 0/0 form. What I am supposed to do here ?

9. May 13, 2015

### Dick

You are given f'(0)=3. What's that as a difference quotient? Halls gave you a hint in post 3.

10. May 13, 2015

### Raghav Gupta

$$f'(0) = \lim_{h \to 0} \frac{f(0)f( h) - f(0)}{h}.$$

Now what? Again 0/0 form.

11. May 13, 2015

### Dick

Think about it some more. The limit is 3. You are given that. As for what f(0) is you were also hinted that.

12. May 13, 2015

### Raghav Gupta

Okay , applying L Hopital rule getting f'(h) = 3
Then using in LH rule once again in f'(5) getting 2f'(h) = 6= f'(5)
Was it supposed to be done by LH rule?

13. May 13, 2015

### Dick

No, you don't need LH. And what is f'(h)=3 supposed to mean? f'(0)=3. You haven't thought about this enough.

14. May 13, 2015

### Raghav Gupta

Not,getting
$$f'(0) =3= \lim_{h \to 0} \frac{f( h) - 1}{h}.$$

15. May 13, 2015

### Dick

Good so far. Now figure out f'(5). It has that expression in it. Don't you see?

16. May 13, 2015

### Raghav Gupta

Yeah f'(5) = 2*3 = 6 by that expression. Thanks Dick.
Thanks to other people also.

17. May 13, 2015

### pasmith

There is a subtle error in the question.

You can prove that $f(x) = f(1)^x = e^{x \ln (f(1))}$ and differentiate that by the chain rule. The function is completely determined by f(1), but you're given two pieces of information with which to try to find it. They are not consistent.

18. May 13, 2015

### Raghav Gupta

How f(x) = f(1)x ?
A counter example-
Putting x = 0
We get f(0) = f(1)
Putting x= 5
We get f(5) = f(1)
But f(5) ≠ f(0)
As we are given f(5) = 2 and we know f(0) = 1

19. May 13, 2015

### SammyS

Staff Emeritus
No, f(0) = (f(1))0 = 1

20. May 13, 2015

### Raghav Gupta

Oh, so it is
(f(1))x not f(1)x
How can we prove f(x) = (f(1))x ?

21. May 13, 2015

### SammyS

Staff Emeritus
Ask Mr. (or Ms.) Smith to be certain.

...but

If f(x) = Cx for some C > 0, then we have f(x+y) = f(x)f(y) .

22. May 13, 2015

### Raghav Gupta

How? Is it some theorem and if it is what it is called?
Can you tell me that Mr. Sammy ?

23. May 13, 2015

### SammyS

Staff Emeritus
Try it. Use some basic algebra.

If I write it out for you, then Mark and other mentors & homework helpers would be displeased with me. (In fact I would then be displeased with me. )

24. May 13, 2015

### Raghav Gupta

Got it, it was basic algebra really.
Wow, is this a paradox( This seems true but contradicts itself) ?
f(x) = (f(0))x
Differentiating,
f'(x) = (f(0))xlogef(0)
Putting x = 0
f'(0) = 0, which is not true since in question it is given f'(0) = 3

25. May 13, 2015

### SammyS

Staff Emeritus
As pasmith said, the conditions given are inconsistent.

Therefore there's an inherent contradiction.