Expert: Paul Klarreich - 5/9/2009

Question
QUESTION: Dear Sir,

I am trying to solve find the derivative of arccos ((1-x)/(1+x)) I can not solve this. Could you please give me a hand and explain me the procedure?

Thank you.

ANSWER: Questioner:   Ignacio
Country:  Spain
Category:  Calculus
Private:  No
 
Subject:  Inverse derivative
Question:  Dear Sir,

I am trying to find the derivative of arccos ((1-x)/(1+x)). Could you please give me a hand and explain me the procedure?

Thank you.
..................................
Hola, Ignacio,

y = arccos ((1-x)/(1+x))

Try the chain rule, with  

y = arccos u

and
   1 - x
u = ------
   1 + x
..................
dy   dy du
-- = -- --
dx   du dx
         -1
dy/du = -------------
       sqrt(1 - u^2)
..................
du/dx =  (using quotient rule)

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

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

   -2
----------
(1+x)^2
....................
Put them together:


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


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


A little simplification: Multiply top&bottom by (1+x)

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


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


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


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

         2
dy/dx = ---------------------
       2 sqrt(x) (1 + x)


         1
dy/dx = -----------------
        sqrt(x) (1 + x)


Whew!
.............................
P.S. You can also try implicit differentiation:

y = arccos ((1-x)/(1+x))
       1 - x  
cos y = -----
       1 + x

Diff:

- sin y  dy/dx = D[(1-x)/(1+x)]
The algebra probably works out the same way.


---------- FOLLOW-UP ----------

QUESTION: Hola Senor,

Gracias for your answer. The derivative of arccos x= -1/(square root(1-x^2)) Could you please explain to me why? Do I have to learn this formula by heart? What is the logic behind it?

Thank you very much.

ANSWER:  Questioner:   Ignacio
Country:  Spain
Category:  Calculus
Private:  No
 
---------- FOLLOW-UP ----------

QUESTION: Hola Senor,

Gracias for your answer. The derivative of arccos x= -1/(square root(1-x^2)) Could you please explain to me why?

>> You prove it by implicit differentiation:

y = arccos(x) means

x = cos y

Differentiate both sides:

1 = - sin y dy/dx
         - 1
dy/dx = ----------   << solve.
        sin y

Now cos^2(y) + sin^2(y) = 1

so  sin y = sqrt(1 - cos^2(y))

and  sin y = sqrt(1 - x^2)

Finally,

         - 1
dy/dx = --------
        sin y

         - 1
dy/dx = --------------
       sqrt(1 - x^2)

................................
Do I have to learn this formula by heart?

>> Yes, you do.

What is the logic behind it?

>> You learn it because it comes up often enough, especially when you do integrals next month.

Thank you very much.


---------- FOLLOW-UP ----------

QUESTION: Dear Sir,

I do not understand this step:

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


A little simplification: Multiply top&bottom by (1+x)

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

Thank you.


Answer
Questioner:   Ignacio
Country:  Spain
Category:  Calculus
Private:  No
 
---------- FOLLOW-UP ----------

QUESTION: Hola Senor,

Gracias for your answer. The derivative of arccos x= -1/(square root(1-x^2)) Could you please explain to me why?

>> You prove it by implicit differentiation:

y = arccos(x) means

x = cos y

Differentiate both sides:

1 = - sin y dy/dx
         - 1
dy/dx = ----------   << solve.
        sin y

Now cos^2(y) + sin^2(y) = 1

so  sin y = sqrt(1 - cos^2(y))

and  sin y = sqrt(1 - x^2)

Finally,

         - 1
dy/dx = --------
        sin y

         - 1
dy/dx = --------------
       sqrt(1 - x^2)

................................
Do I have to learn this formula by heart?

>> Yes, you do.

What is the logic behind it?

>> You learn it because it comes up often enough, especially when you do integrals next month.

Thank you very much.
==========================

---------- FOLLOW-UP ----------

QUESTION: Dear Sir,

I do not understand this step:

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


A little simplification: Multiply top&bottom by (1+x)

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

Thank you.
...................................
Hola, Ignacio,

Yes, I thought you might have a bit of trouble with that step.  I agree it is not the easiest part.  Here it is again, with a WARNING:

USE COURIER FONT TO VIEW IT, OR IT WILL NOT MAKE SENSE.

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


Let's combine the terms inside the radical sign:

    (1-x)^2
1 - --------    << Lowest common denominator is (1+x)^2
    (1+x)^2

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

Put them back now:

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

Take root on top and bottom:

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


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


Now a factor of (1+x) cancels:

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

And now you can go on from there.  The key step is realizing that (1+x)^2 inside the root  becomes (1+x) outside the root. (And I apologize for such 'non-mathematical' language.)

I hope this helps.  And I agree that the algebra was somewhat unclear.