# Differentiate hyperbolic function

1. Dec 7, 2011

### sharks

The problem statement, all variables and given/known data
If lny = sinh^(-1)(x), prove that

(1+ x^2)y'' + xy' - y = 0

The attempt at a solution

I have tried various (unsuccessful) ways of doing this, but the basic procedure that i've done is:

D.w.r.t.x for lny = sinh^(-1)(x)

This gives: (1/y)y' = 1/(1 + x^2)

To obtain y'', d.w.r.t.x for equation above.

This is the part where i'm not sure if i should use product rule for differentiating (1/y)y'

Is is correct?
-y^(-2)y' + y^(-1)y''

Or is this the right way of doing it?
-y^(-2)y''

2. Dec 7, 2011

### LCKurtz

You must use the product (or quotient) rule.

3. Dec 7, 2011

### Quinzio

Well, you should be aware that:
$sinh^{-1}x=log(x+\sqrt{x^2+1})$

so you have
$\log y=\log(x+\sqrt{x^2+1})$

from now on it should be simple.

4. Dec 7, 2011

### vela

Staff Emeritus
You differentiate sinh-1 x incorrectly. You should have
$$\frac{y'}{y} = \frac{1}{\sqrt{1+x^2}}$$Then I'd get rid of the quotients to make things simpler:
$$y'\sqrt{1+x^2} = y$$before differentiating. (I just never really liked the quotient rule.)

Last edited: Dec 7, 2011
5. Dec 8, 2011

### sharks

I have taken into account all of your advices. OK, now here's the evolution of this problem from my side.

So, making y' the subject of formula, i get y' = y/[(1 + x^2)^(1/2)]

I further differentiate w.r.t.x the equation by vela above (i apologize for not using the better-looking LaTex, but i still have a lot to learn on how to use it):

So, y''(1 + x^2)^(1/2) + y'/[2(1 + x^2)^(1/2)] = y'

Now, i make y'' the subject of formula.

y'' = {y' - y'/[2(1 + x^2)^(1/2)]}/[(1 + x^2)^(1/2)]

Then, i replace y' and y'' above, into the original equation: (1+ x^2)y'' + xy' - y, hoping that it will all reduce to 0. After substituting, i will end up with a (rather long) equation involving x, y and y' (which i have to substitute again). Is this the right direction to solve this problem?

6. Dec 8, 2011

### vela

Staff Emeritus
You need to use the chain rule when calculating the second term.
You're making more work for yourself than you need to.

Hint: After you differentiated, the y'' term almost looks like what you want.

7. Dec 8, 2011

### sharks

OK, i got it. Thank you all very much.