How do you differentiate #f(x)=1/cos(e^arccos(lnx))# using the chain rule?

1 Answer
Mar 30, 2018

Answer:

#color(blue)(f'(x) = - tan (e^arccos(ln x))* sec (e^arccos(ln x))* e^arccos(ln x)) /color(blue)( x * sqrt(1 - ln^2 x))#

Explanation:

#f(x) = 1 / cos (e^arccos(ln x))#

#f'(x) = (d/(dx)cos(e^arccos(ln x)) ) / cos ^2 (e^arccos (ln x))#

(Multiply and divide by #cos(e^arccos(ln x))#

http://faculty.wlc.edu/buelow/Web%20Library/chain_rule.htm

Applying chain rule,

#=> (- sin (e^arccos(ln x)) *( (d/(dx)) (e^arccos (ln x)))) / cos^2 (e^arccos(ln x))#

#=> (-sin (e^arccos(ln x)) * e^arccos(ln x) * (d/(dx)) arccos (ln x)) / cos^2 (e^arccos(ln x))#

#=> (-1/sqrt(1 - ln^2 x) * (d/(dx))(ln x) * sin (e^arccos(ln x)) * e^arccos(ln x)) / cos^2 (e^arccos(ln x))#

#=>- (sin (e^arccos(ln x)) * e^arccos(ln x)) /( x * sqrt(1 - ln^2 x)* cos^2 (e^arccos(ln x))#

#=>- (tan (e^arccos(ln x))* (sec (e^arccos(ln x))* e^arccos(ln x))) /( x * sqrt(1 - ln^2 x))#