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

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Answer 1 · 2016-08-27T18:39:07

df(x) =#-sec^2(x^2+1)*cos(x^2+1)*2x-2tanx*cos^2x#

The decision based on the rules of differentiation of a composite function

Answer 2 · 2016-09-04T09:01:09

#2x sec(x^(2) + 1) tan(x^(2) + 1) - 2 sec^(2)(x) tan(x)#

We have: #f(x) = sec(x^(2) + 1) - tan^(2)(x)#

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

This function can be differentiated using the "chain rule" and the "sum rule".

Let #u = x^(2) + 1 => u' = 2x# and #v = sec(u) => v' = sec(u) tan(u)#:

#=> f'(x) = 2x cdot sec(u) tan(u) - (d) / (dx) (tan^(2)(x))#

#=> f'(x) = 2x sec(u) tan(u) - (d) / (dx) (tan^(2)(x))#

We can now replace #u# with #x^(2) + 1#:

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

Now, let #u = tan(x) => u' = sec^(2)(x)# and #v = u^(2) => v' = 2 u#:

#=> f'(x) = 2x sec(x^(2) + 1) tan(x^(2) + 1) - (sec^(2)(x) cdot (2 u))#

#=> f'(x) = 2x sec(x^(2) + 1) tan(x^(2) + 1) - (2 sec^(2)(x) u)#

We can now replace #u# with #tan(x)#:

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

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