# Homework Help: Find f(x) from following data

1. May 30, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
If f(x/y) = f(x)/f(y) f(y)≠0 and f'(1)=2 find f(x).

2. Relevant equations

3. The attempt at a solution
Diff both sides wrt x
f'(x/y)*1/y=f'(x)/f(y)
putting x=1
f'(1/y)*1/y=2/f(y)

2. May 30, 2013

### Saitama

Your first step is wrong. Apply chain rule.

3. May 30, 2013

### utkarshakash

I have differentiated wrt x treating y as a constant.

4. May 30, 2013

### tiny-tim

hi utkarshakash!
ok so far

now you need a relation between f(1/y) and f(y)

5. May 30, 2013

### Saitama

Ah yes, sorry about that. Now as tiny-tim said, a relation between f(y) and f(1/y) would be useful.

Hint: Put x=1 in the original equation.

6. May 30, 2013

### haruspex

You might find the path a little more obvious if you differentiate wrt y instead. Note that you can easily deduce f(1).

7. May 31, 2013

### utkarshakash

How to find that?

8. May 31, 2013

### tiny-tim

apply the formula in the question to f(1/y)

9. May 31, 2013

### AGNuke

If I had to guess, I'd say $f(x)=x^2$.

10. May 31, 2013

### utkarshakash

Yes the answer is correct.

11. May 31, 2013

### tiny-tim

ok, now derive it!

12. May 31, 2013

### utkarshakash

f(1)=f(y)f(1/y)

13. May 31, 2013

### tiny-tim

ok, now combine that with your original equation …
… to get a differential equation in f(1/y)

(also, isn't it fairly obvious what f(1) is?)

14. May 31, 2013

### Vibhor

Please can you help me understand a very important concept and i.e why have we differentiated wrt x treating y as a constant .

Isnt x and y both variable ? Why is applying chain rule to differentiate wrong as earlier suggested by Pranav-Arora ?

Thanks

15. May 31, 2013

### tiny-tim

yes, x and y are independent variables, and so we can ∂/∂x keeping y constant, or ∂/∂y keeping x constant
i think you've misread that …

everybody's been applying the chain rule

16. May 31, 2013

### Saitama

I have been trying this question from quite some time and I always ended up with the same differential equation you are talking about but I don't see a way to solve it.

What I end up with is this:
$$\frac{∂f(1/y)}{∂x}\cdot \frac{1}{y}=2f(1/y)$$
I don't see how can i solve this but if I replace y with 1/x, I do get the right answer but is it valid to replace y with 1/x?

Thanks!

17. May 31, 2013

### Vibhor

Thanks tiny-tim for clarification...Yes I indeed misread things

18. May 31, 2013

### utkarshakash

f'(1/y)/f(1/y)=2y (Assuming x=1)

But I haven't learned how to solve DE's.

19. May 31, 2013

### utkarshakash

Yes of course it is valid. Since the function is valid for all y belonging to ℝ.
But can you tell me the method to solve this equation? I haven't been taught DE yet.

20. May 31, 2013

### Saitama

I still don't get this.

Assuming that replacing y with 1/x is valid,
$$\frac{∂f(x)}{∂x}\cdot x=2f(x)$$
Rearranging,
$$\frac{∂f(x)}{f(x)}=2\frac{∂x}{x}$$
You do know how to solve this now. ($\int dx/x=\ln x+c$)

21. May 31, 2013

### Mute

The equation you've written here isn't correct. The derivative is with respect to the entire argument of f, not just x, because of the chain rule:

$$\frac{\partial}{\partial x} f(x/y) = \left.\frac{df}{dz}\right|_{z = x/y} \frac{\partial(x/y)}{\partial x} = \left.\frac{df}{dz}\right|_{z = x/y} \frac{1}{y}.$$

You can then set x = 1 to get $\left.\frac{df}{dz}\right|_{z = 1/y} \frac{1}{y}$ on the left hand side. Because the derivative is with respect to the argument of f, and not just x or y, you can legitimately change variables and it doesn't affect the $\left.\frac{df}{dz}\right|_{z = 1/y}$ term except that you're replacing 1/y with x.

Of course, if everyone were to just follow haruspex's advice the solution would come about much more quickly:

22. May 31, 2013

### tiny-tim

no, f'(1/y) is ∂f(1/y)/∂(1/y)

23. May 31, 2013

### tiny-tim

the first thing is to write 1/y = z, to make things easier!

f'(z)/f(z) = 2/z

then

f'(z)/f(z) dz = 2/z dz

now integrate both sides

24. May 31, 2013

### Saitama

But OP said he differentiated the equation w.r.t x.

25. May 31, 2013

### tiny-tim

let's see …
he used the chain rule …

∂/∂x f(x/y)

= ∂/∂(x/y) f(x/y) * ∂(x/y) ∂x

= ∂/∂(x/y) f(x/y) * (1/y)

… what rule did you use?

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