What is the derivative of #y=sec^3(x)#?

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Answer 1 · 2014-07-25T21:18:23

Answer is

#y' = 3* sec^2x*secx*tanx#

The solution is,

For problems like these, #y=f(x)^n#

then #y' = n*f(x)^(n-1)*f'(x)# (this is the Power Chain Rule)

Similarly for the question asked above

#y' = 3* sec^2x*secx*tanx#