What is the derivative of #y=4 sec ^2(x)#?

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Answer 1 · 2014-08-15T13:58:57

#y'=8sec^2(x)tan(x)#

Explanation :

let's start with general function, #y=(f(x))^2#

differentiating with respect to #x# Using Chain Rule,

#y'=2*f(x)*f'(x)#

Similarly following for given problem, yields

#y= 4*sec^2(x)#

#y'=4*2*sec(x)*sec(x)tan(x)#

#y'=8sec^2(x)tan(x)#