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 $tan(x)$ :

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

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

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