What is the second derivative of the function #f(x)=sec x#?

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Answer 1 · 2018-07-16T10:25:40

#f''(x)=sec x(\sec^2 x+\tan^2 x)#

given function:

#f(x)=\sec x#

Differentiating w.r.t. #x# as follows

#\frac{d}{dx}f(x)=\frac{d}{dx}(\sec x)#

#f'(x)=\sec x\tan x#

Again, differentiating #f'(x)# w.r.t. #x#, we get

#\frac{d}{dx}f'(x)=\frac{d}{dx}(\sec x\tan x)#

#f''(x)=\sec x\frac{d}{dx}\tan x+\tan x\frac{d}{dx}\secx#

#=\sec xsec^2 x+\tan x\sec x\tan x#

#=sec^3 x+\sec x\tan^2 x#

#=sec x(\sec^2 x+\tan^2 x)#