How do you differentiate # f(x) = sec(x^2 + 1)^2-tanx #?

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Answer 1 · 2018-07-06T16:30:43

#f'(x)=4x(x^2+1)*sec(x^2+1)^2tan(x^2+1)^2-sec^2x#

We know that ,

#(1)d/(d theta)(sectheta)=secthetatantheta#

#(2)d/(d theta)(tantheta)=sec^2theta#

Here ,

#f(x)=sec(x^2+1)^2-tanx#

Diff.w.r.t. #x# using (1) ,(2) and chain rule ,

#f'(x)=sec(x^2+1)^2tan(x^2+1)^2(d/(dx)(x^2+1)^2)-sec^2x#

#f'(x)=sec(x^2+1)^2tan(x^2+1)^2[2(x^2+1)2x]-sec^2x#

#f'(x)=4x(x^2+1)*sec(x^2+1)^2tan(x^2+1)^2-sec^2x#
………………………………………………………………………………..
Note :

#(i)sec(x^2+1)^2 !=sec^2(x^2+1)#

#(ii)sec(x^2+1)^2=sec(x^4+2x^2+1)#