Expert: Paul Klarreich - 10/10/2006

Question
My name is David and Im taking Calculus I right now.

A) Find d/dx of sqaure root of (169/x)- x.
I started the problem by factoring out (1/x) and got square root of (1/x)(169-x^2). Im tried to use the quotient rule but the alegbra was too messy to work. Now, im stuck.

B)Find the square root of ( (x^2 +1)^2 + square root of 1 + (x^2 +1)^2.)
I figure I have to use the chain rule for this:
F(u)= square root of u, U(x)= (x^2+1)^2 +square root of 1+ (x^2+1)^2. The alegbra is just too messy.  

Answer
David S. Asks in Category Calculus ...
 
Subject:  Finding derivative
Private:  no
 
Question:  My name is David and Im taking Calculus I right now.

A) Find d/dx of sqaure root of (169/x)- x.
I started the problem by factoring out (1/x) and got square root of (1/x)(169-x^2). Im tried to use the quotient rule but the alegbra was too messy to work. Now, im stuck.

B)Find the square root of ( (x^2 +1)^2 + square root of 1 + (x^2 +1)^2.)
I figure I have to use the chain rule for this:
F(u)= square root of u, U(x)= (x^2+1)^2 +square root of 1+ (x^2+1)^2. The alegbra is just too messy
..........................................
Hi, David,

Yes, the algebra in calculus I can be very 'messy'.  

[Every fall, some rookie quarterback comes into an NFL game for the first time.  The defensive linemen all pile on top of him, grinding him into the turf, saying, sweetly, "Welcome to the NFL."]

So, welcome to calculus.


However, there may be approaches to an example that may reduce the 'mess.'  And even if it's going to be messy, the solution is to break up the example into two or more pieces, work on them one at a time, then put the stuff together at the end.

WARNING: THE FOLLOWING DISCUSSION MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

A) My suggestion for this one is NOT to produce a quotient, but write:

y = sqrt(169x^-1 - x)

Then organize the chain rule:
                                         1
y = u^1/2,  and  
                         1
dy/du = 1/2 u^-1/2 =  ---------
                     2 sqrt(u)

u = 169x^-1 - x, and

du/dx = - 169x^-2 - 1 =
-169       - 169 - x^2
---- - 1 = -----------
x^2           x^2

Now put them together:

dy    - 169 - x^2
-- = ---------------------
dx   2 x^2 sqrt(169/x - x)

And that's it.


B)Find the square root of ( (x^2 +1)^2 + square root of 1 + (x^2 +1)^2.)

I assume you mean find the derivative of:

sqrt( (x^2 + 1)^2  + sqrt(1 + (x^2 + 1)^2 ) )

Are you sure you parenthesized correctly?

Same stuff:

y = sqrt(u) and  u = that stuff.
           1
dy/du = ---------, which you'll rewrite later.
       2 sqrt(u)

u = (x^2 + 1)^2  + sqrt(1 + (x^2 + 1)^2 ), which I suggest you expand first.

u = x^4 + 2x^2 + 1 + sqrt(1 + x^4 + 2x^2 + 1)

u = x^4 + 2x^2 + 1 + sqrt(x^4 + 2x^2 + 2)

Now do the sqrt part:

v = sqrt(z), and z = x^4 + 2x^2 + 2

dv       1
-- = ---------
dz   2 sqrt(z)

dz
-- = 4x^3 + 4x
dx

So

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


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


Just put them together now:

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

P.S. There are some symbolic algebra programs out there that can produce the derivative.  [It is, after all, a very mechanical process.]  I tried one of these, and the answer it produced was:

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

NOW THAT'S WHAT I CALL MESSY.