+0  
 
0
348
1
avatar

\(\frac{d}{dx} (ln|cos\frac{pi}{x}|)\)

So far I've been able to go through some of the work

The first chain rule:

(1/x) * (abs(cos(pi/x))) * (x/(abs(x))) * (Cos(pi/x)) * [ -sin(x) * (pi/x) * { ( x(0) - (pi)(1) )/ (x^2) } ]

that eventually simplifies to:

( (cos(pi/x)) * (-sin(x)) * (pi^2) * (Abs( Cos(pi/x) )) ) / ( (x^3) * (abs(x)) )

 

(I'm not going to include all my simplification work, because it took me 15 minutes to type what I already have)

 

I'm relatively certain this work is correct - my problem is that my answer choices arent even remotely similar, so I'm missing something. either I differentiated wrong, or im missing some way to simplify further.

 

Answer Choices:

A. (-pi) / ( x^2 * cos(pi/x) )

 

B. -tan(pi/x)

 

C. 1 / Cos(pi/x)

 

D. (pi/x) * tan(pi/x)

 

E. (pi/x^2) * tan(pi/x)

 

I have absolutely no clue where to go from what I currently have... Please help

 Oct 27, 2020
 #1
avatar
0

Possible derivation:
d/dx(log(abs(cos(π/x))))

Using the chain rule, d/dx(log(abs(cos(π/x)))) = ( dlog(u))/( du) ( du)/( dx), where u = abs(cos(π/x)) and d/( du)(log(u)) = 1/u:
 = (d/dx(abs(cos(π/x))))/abs(cos(π/x))

Using the chain rule, d/dx(abs(cos(π/x))) = ( dabs(u))/( du) ( du)/( dx), where u = cos(π/x) and d/( du)(abs(u)) = u/abs(u):
 = ((cos(π/x) d/dx(cos(π/x)))/(abs(cos(π/x))))/abs(cos(π/x))

Simplify the expression:
 = (cos(π/x) (d/dx(cos(π/x))))/abs(cos(π/x))^2

Using the chain rule, d/dx(cos(π/x)) = ( dcos(u))/( du) ( du)/( dx), where u = π/x and d/( du)(cos(u)) = -sin(u):
 = -d/dx(π/x) sin(π/x) cos(π/x)/abs(cos(π/x))^2

Factor out constants:
 = -(cos(π/x) sin(π/x))/abs(cos(π/x))^2 π d/dx(1/x)

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = -1.
d/dx(1/x) = d/dx(x^(-1)) = -x^(-2):
 = -(π cos(π/x) sin(π/x))/abs(cos(π/x))^2 (-1)/(x^2)

Simplify the expression:
 = (π cos(π/x) sin(π/x))/(x^2 abs(cos(π/x))^2)

Simplifying powers, abs(cos(π/x))^2 = cos^2(π/x):


= (π tan(π/x))/x^2     [Courtesy of Mathematica 11 Home Edition]

 Oct 27, 2020

1 Online Users

avatar