How do you find the derivative of #f(x) = arctan(cos(3x))#?

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Answer 1 · 2017-05-29T10:48:54

#=-(3sin(3x) )/(1+cos^2(3x))#

#d/dx(tan^-1(cos(3x)))#

Step 1. Use the chain rule
#d/dx(tan^-1(cos(3x)))=(d(tan^-1(u)))/(du)(du)/(dx)#,

where #u=cos(3x)# and #d/(du)(tan^-1(u))=1/(1+u^2)#

#=(d/dx(cos(3x)))/(1+cos^2(3x))#

Step 2. Using the chain rule again,

#d/dx(cos(3x))=(d(cos(u)))/(du)(du)/(dx)#,

where #u=3x# and #d/(du)(cos(u))-sin(u)#, gives

#=(-d/dx(3x)sin(3x) )/(1+cos^2(3x))#

Step 3. Factor out constants

#=-3(sin(3x) )/(1+cos^2(3x))d/dx(x)#

ANSWER: #=-(3sin(3x) )/(1+cos^2(3x))#