What is the derivative of #cos(tan x)#?

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Answer 1 · 2016-06-23T00:06:36

# - sin (tan(x)) * sec^2 x#

#y = cos (tan x)#

#= cos (p), \qquad p = tan(x)#

so you chain rule it!

#dy/(dp) = - sin (p)#

#(dp)/dx = sec^2 x#

# dy/dx = dy/(dp) * (dp)/dx #

# = - sin (p) * sec^2 x#

# = - sin (tan(x)) * sec^2 x#