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

Question

1868 views around the world

You can reuse this answer
Creative Commons License

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 :

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

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

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