If y = e^ 1 2log(1+ tan^2 x), then d^2y dx^2 is equal to:

The domain consists of all x-values that can be input into the function.

For $\cos^{-1}$cos−1 (inverse cosine), the input must be between $-1$−1 and $1$1, inclusive.

---

Since the function is $y = \cos^{-1}(x^2 - 4)$y=cos−1(x2−4), the expression $(x^2 - 4)$(x2−4) must satisfy:

$$-1 \leq x^2 - 4 \leq 1$$−1≤x2−4≤1

---

For the left inequality:

$$x^2 - 4 \geq -1$$x2−4≥−1

$$x^2 \geq 3$$x2≥3

$$|x| \geq \sqrt{3}$$∣x∣≥3​

This gives: $x \leq -\sqrt{3}$x≤−3​ or $x \geq \sqrt{3}$x≥3​

---

For the right inequality:

$$x^2 - 4 \leq 1$$x2−4≤1

$$x^2 \leq 5$$x2≤5

$$|x| \leq \sqrt{5}$$∣x∣≤5​

This gives: $-\sqrt{5} \leq x \leq \sqrt{5}$−5​≤x≤5​

---

Both conditions must be satisfied simultaneously.

On the left side: $x \leq -\sqrt{3}$x≤−3​ and $x \geq -\sqrt{5}$x≥−5​ gives $[-\sqrt{5}, -\sqrt{3}]$[−5​,−3​]

On the right side: $x \geq \sqrt{3}$x≥3​ and $x \leq \sqrt{5}$x≤5​ gives $[\sqrt{3}, \sqrt{5}]$[3​,5​]

---

Domain: $[-\sqrt{5}, -\sqrt{3}] \cup [\sqrt{3}, \sqrt{5}]$[−5​,−3​]∪[3​,5​]