1. Problem: Izračunajte drugu derivaciju funkcije $$f(x) = \sinh\left(\frac{1}{x^2 - 4}\right)$$. 2. Formula: Koristimo lančanu i pravilo za derivaciju hiperboličkog sinusa: $$\frac{d}{dx} \sinh(u) = \cosh(u) \cdot u'$$ 3. Prvo izračunajmo prvu derivaciju $f'(x)$: Neka $$u = \frac{1}{x^2 - 4}$$ Tada je $$u' = \frac{d}{dx} \left(\frac{1}{x^2 - 4}\right) = \frac{d}{dx} (x^2 - 4)^{-1} = -1 \cdot (x^2 - 4)^{-2} \cdot 2x = -\frac{2x}{(x^2 - 4)^2}$$ Dakle, $$f'(x) = \cosh\left(\frac{1}{x^2 - 4}\right) \cdot \left(-\frac{2x}{(x^2 - 4)^2}\right) = -\frac{2x \cosh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^2}$$ 4. Sada izračunajmo drugu derivaciju $f''(x)$ koristeći pravilo produkta: $$f''(x) = \frac{d}{dx} \left(-\frac{2x \cosh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^2}\right)$$ Neka $$g(x) = -2x, \quad h(x) = \frac{\cosh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^2}$$ Tada $$f''(x) = g'(x) h(x) + g(x) h'(x)$$ Izračunajmo: $$g'(x) = -2$$ Za $h(x)$ koristimo pravilo produkta i lančanu: $$h(x) = \cosh\left(u\right) \cdot (x^2 - 4)^{-2}$$ $$h'(x) = \cosh'(u) \cdot u' \cdot (x^2 - 4)^{-2} + \cosh(u) \cdot \frac{d}{dx} (x^2 - 4)^{-2}$$ Znamo da je $$\frac{d}{dx} \cosh(u) = \sinh(u) \cdot u'$$ Dakle, $$h'(x) = \sinh(u) \cdot u' \cdot (x^2 - 4)^{-2} + \cosh(u) \cdot (-2) (x^2 - 4)^{-3} \cdot 2x$$ Zamenjujemo $u' = -\frac{2x}{(x^2 - 4)^2}$: $$h'(x) = \sinh\left(\frac{1}{x^2 - 4}\right) \cdot \left(-\frac{2x}{(x^2 - 4)^2}\right) \cdot (x^2 - 4)^{-2} - 4x \cosh\left(\frac{1}{x^2 - 4}\right) (x^2 - 4)^{-3}$$ Pojednostavimo: $$h'(x) = -\frac{2x \sinh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^4} - \frac{4x \cosh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^3}$$ 5. Konačno, $$f''(x) = (-2) \cdot \frac{\cosh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^2} + (-2x) \cdot \left(-\frac{2x \sinh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^4} - \frac{4x \cosh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^3}\right)$$ Razvijamo: $$f''(x) = -\frac{2 \cosh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^2} + \frac{4x^2 \sinh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^4} + \frac{8x^2 \cosh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^3}$$ To je druga derivacija funkcije $f(x)$. **Finalni odgovor:** $$f''(x) = -\frac{2 \cosh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^2} + \frac{4x^2 \sinh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^4} + \frac{8x^2 \cosh\left(\frac{1}{x^2 - 4}\right)}{(x^2 - 4)^3}$$