Expert: Paul Klarreich - 10/24/2006

Question

Hi Paul,

The way the sum is written is as i typed it.
y = ln e^-x SQRT[(1+2x)/(1-2x)]

Maybe there should be brackets arboud the exponential function so that its all under ln but this is not the way its shown. Above is the way its shown in the text book.

Should i try it the other way and see will it work out?


-------------------------------------------
The text above is a follow-up to ...

-----Question-----
Hi Paul,

I have the following question, and i just want a hint on how to start it off.

If y = ln e^-x SQRT[(1+2x)/(1-2x)]

show that dy/dx = (1+4x^2)/(1-4x^2)

I know that ln e^-x = -x

and

SQRT[(1+2x)/(1-2x)]

equals [(1+2x)/(1-2x)]^1/2

so should i treat the sum as a product and differentiate using the product rule?

ie. -x [(1+2x)/(1-2x)]^1/2

Please advise if you can, before i go any further.

Many thanks,
Annie



-----Answer-----
Questioner:  Annie
Category:  Advanced Math
Private:  no
 
Subject:  Differentiating Natural Logs
Question:  Hi Paul,

I have the following question, and i just want a hint on how to start it off.

If y = ln e^-x SQRT[(1+2x)/(1-2x)]

show that dy/dx = (1+4x^2)/(1-4x^2)

I know that ln e^-x = -x

and

SQRT[(1+2x)/(1-2x)]

equals [(1+2x)/(1-2x)]^1/2

so should i treat the sum as a product and differentiate using the product rule?

ie. -x [(1+2x)/(1-2x)]^1/2

Please advise if you can, before i go any further.

Many thanks,
Annie
......................................
Hi, Annie,

I think there must be some error in the way you are parenthesizing the expressions.  If you are differentiating:

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

Then, yes, you need:
The product rule, because this is -x V.
The chain rule, because V = sqrt(U) or U^1/2
The quotient rule, because U is a quotient.

I tried working this out, but I don't get the answer you seem to require:

(1+4x^2)/(1-4x^2)

Are you sure it wasn't supposed to be:

y = ln [ e^-x SQRT[(1+2x)/(1-2x)]  ]

which is a different thing altogether.  Let me know before I do any more on it.  

Answer
Questioner:  Annie
Category:  Advanced Math
 
Subject:  Differentiating Natural Logs
Question:  
Hi Paul,

The way the sum is written is as i typed it.
y = ln e^-x SQRT[(1+2x)/(1-2x)]

Maybe there should be brackets arboud the exponential function so that its all under ln but this is not the way its shown. Above is the way its shown in the text book.

...................................
Hi, Annie,

Since  ln e^-x is just a funny way to write  -x, it seems strange for the example to be written that way.  It's just writing:

y = - x sqrt[(1+2x)/(1-2x)]

If this is the case, you can avoid one of the complications by writing:

y = - sqrt[x^2 (1+2x)/(1-2x)] = - sqrt[(x^2 + 2x^3)/(1-2x)]

Then you just need the chain and quotient rules.  Let's try that:

******** USUAL WARNING ABOUT FIXED-SIZE FONTS. ********


y = - sqrt(u) = - u^(1/2),  and dy/du = -1/2u^(-1/2)
        x^2 + 2x^3     
And  u = ----------
        1 - 2x

du   (1 - 2x)(2x + 6x^2) - (x^2 + 2x^3)(-2)
-- = ---------------------------------------
dx          (1 - 2x)^2


du   2x(1 - 2x)(1 + 3x) + 2x^2 + 4x^3
-- = ---------------------------------------
dx          (1 - 2x)^2

du   2x[1 + x - 6x^2] + 2x^2 + 4x^3
-- = ---------------------------------
dx          (1 - 2x)^2

du   2x + 2x^2 - 12x^3 + 2x^2 + 4x^3
-- = ---------------------------------
dx          (1 - 2x)^2

du   2x + 4x^2 - 8x^3
-- = ------------------
dx        (1 - 2x)^2

du   2x(1 +2x - 4x^2)
-- = ----------------
dx      (1 - 2x)^2

Now

dy      
-- = -1/2 u^(-1/2)
du   

dy      
-- = -1/2 u^(-1/2)
du   

dy   -1 sqrt(1-2x)
-- = -------------------  << Note inversion trick.
du    2 sqrt(x^2 + 2x^3)

Put them together:

dy   -1 sqrt(1-2x) 2x(1 +2x - 4x^2)
-- = ------------------------------
dx    2 sqrt(x^2 + 2x^3) (1 - 2x)^2

dy   -1 sqrt(1 - 2x) x(1 +2x - 4x^2)
-- = ------------------------------
dx    x sqrt(1 + 2x) (1 - 2x)^2

dy   -1 sqrt(1 - 2x) (1 +2x - 4x^2)
-- = ------------------------------
dx      sqrt(1 + 2x) (1 - 2x)^2


dy    -1 (1 +2x - 4x^2)
-- = ------------------------------
dx    sqrt(1 + 2x)(1 - 2x) sqrt(1 - 2x)


dy    -1 (1 +2x - 4x^2)
-- = -----------------------
dx    sqrt(1 - 4x^2)(1 - 2x)

That's about as far as it goes.

Note: There is a lot of simplification here; in my declining years it has become possible for me to make an error. (!)  In fact, I made one in March of 2005.  So I decided to check it out.  I found a symbolic algebra progam called Maxima, at Sourceforge.net.  [Free download.]

It takes a bit of getting used to, but I managed to plug your function into it and got this answer.  It might not help you, but I feel a lot better.