How do you take the derivative of #cos(tan x)#?

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Answer 1 · 2018-04-18T00:32:00

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

Use the chain rule, which states that,

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

Let #u=tan(x),:.(du)/dx=sec^2(x)#

Then #y=cos(u),:.dy/(du)=-sin(u)#

Combining, we get:

#dy/dx=-sin(u)*sec^2(x)#

#=-sin(u)sec^2(x)#

Substitute back #u=tan(x)# to get the final answer:

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