Expert: Paul Klarreich - 10/13/2007

Question
i can't seem to get the differential of the formula  y= x/2 (cos lnx + sin lnx), i
started by using the chain rule, but the working just looks wrong and so far
from the correct answer. help, please?

Answer
Questioner:   ZEE
Category:  Calculus
Private:  No
 
Subject:  calculus
Question:  I can't seem to get the differential of the formula  y= x/2 (cos lnx + sin lnx), I started by using the chain rule, but the working just looks wrong and so far from the correct answer. help, please?

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Hi, Zee,

This is a product rule example.  (You will use the chain rule for parts of it, too.)

I am going to skip the  1/2  factor for now.  I'll put it back at the end -- saves typing.

y= x(cos lnx + sin lnx)

For the individual 'pieces',  cos(ln x) and sin(ln x), we'll use the chain rule and then put everything together at the end.

D(cos (ln x)) =  - sin(ln x)(1/x) = - (1/x) sin(ln x)

D(sin (ln x)) =    cos(ln x)(1/x) =   (1/x) cos(ln x)

Now product rule for the whole thing:

dy/dx = (x)(- (1/x) sin(ln x) + (1/x) cos(ln x)) + (1)(cos(ln x) + sin(ln x))

dy/dx = - sin(ln x) +  cos(ln x)) + cos(ln x) + sin(ln x)

dy/dx = 2 cos(ln x))

Oh, yes -- that factor of 1/2 makes it:

dy/dx = cos(ln x))

Does that look right?