How do you differentiate #arccos(tan(1/x) )# using the chain rule?

1 Answer
Jan 11, 2017

Answer:

#d/(dx) (arccos(tan(1/x))) = 1/(x^2cos(1/x)sqrt(cos^2(1/x)-sin^2(1/x) #

Explanation:

Let's name:

#y = 1/x#

#u = tan y#

#v= arccos u#

we have that:

#(dv)/(dx) = (dv)/(du)* (du)/(dx) = -1/sqrt(1-u^2) (du)/(dx)#

# (du)/(dx) = (du)/(dy)* (dy)/(dx) = 1/cos^2y (dy)/(dx)#

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

And substituting all the intermediate expressions as functions of #x#:

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