Finding f(x) from Given Data and Derivative: A Scientific Approach

[utkarshakash]

Homework Statement

If f(x/y) = f(x)/f(y) f(y)≠0 and f'(1)=2 find f(x).

Homework Equations

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)

 

[Saitama]

Homework Statement

If f(x/y) = f(x)/f(y) f(y)≠0 and f'(1)=2 find f(x).

Homework Equations

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)

Your first step is wrong. Apply chain rule.

 

[utkarshakash]

Your first step is wrong. Apply chain rule.

I have differentiated wrt x treating y as a constant.

 

[tiny-tim]

hi utkarshakash!

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)

ok so far

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

 

[Saitama]

I have differentiated wrt x treating y as a constant.

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.

 

[haruspex]

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

 

[utkarshakash]

hi utkarshakash!


ok so far

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

How to find that?

 

[tiny-tim]

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

 

[AGNuke]

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

 

[utkarshakash]

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

Yes the answer is correct.

 

[tiny-tim]

ok, now derive it!

 

[utkarshakash]

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

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

 

[tiny-tim]

ok, now combine that with your original equation …

f'(1/y)*1/y=2/f(y)

… to get a differential equation in f(1/y)

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

 

[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

 

[tiny-tim]

Isnt x and y both variable ?

yes, x and y are independent variables, and so we can ∂/∂x keeping y constant, or ∂/∂y keeping x constant

Why is applying chain rule to differentiate wrong as earlier suggested by Pranav-Arora ?

i think you've misread that …

everybody's been applying the chain rule

 

[Saitama]

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

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!

 

[Vibhor]

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

 

[utkarshakash]

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?)

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

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

 

[utkarshakash]

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!

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.

 

[Saitama]

Yes of course it is valid. Since the function is valid for all y belonging to ℝ.

I still don't get this.

But can you tell me the method to solve this equation? I haven't been taught DE yet.

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. ($\int dx/x=\ln x+c$)

 

[Mute]

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!

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:

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

 

[tiny-tim]

f'(1/y)*1/y=2/f(y)

What I end up with is this:
\frac{∂f(1/y)}{∂x}\cdot \frac{1}{y}=2f(1/y)

no, f'(1/y) is ∂f(1/y)/∂(1/y)

 

[tiny-tim]

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

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

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 ​

 

[Saitama]

no, f'(1/y) is ∂f(1/y)/∂(1/y)

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

 

[tiny-tim]

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

let's see …

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)

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?

 

[Saitama]

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?

Thank you tiny-tim, that helped a lot!