SULIT 15. (a) (i) Cari \( f^{\prime}(x) \) bagi \( f(x)=3 x^{2}+1 \) dengan menggunakan pembezaan dengan prinsip pertama. Find \( f^{\prime}(x) \) for \( f(x)=3 x^{2}+1 \) using differentiation by the first principles. (ii) Cari terbitan pertama bagi fungsi \( f(x)=\frac{1}{2} x^{2}-3 x \). Find the first derivative of the function \( f(x)=\frac{1}{2} x^{2}-3 x \).
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Untuk mencari \( f^{\prime}(x) \) menggunakan prinsip pertama, kita akan menggunakan definisi pembezaan: \[ f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Bagi \( f(x) = 3x^{2} + 1 \): \[ f(x+h) = 3(x+h)^{2} + 1 = 3(x^{2} + 2xh + h^{2}) + 1 = 3x^{2} + 6xh + 3h^{2} + 1 \] Kemudian, \[ f^{\prime}(x) = \lim_{h \to 0} \frac{(3x^{2} + 6xh + 3h^{2} + 1) - (3x^{2} + 1)}{h} = \lim_{h \to 0} \frac{6xh + 3h^{2}}{h} = \lim_{h \to 0} (6x + 3h) = 6x \] Oleh itu, \( f^{\prime}(x) = 6x \). Untuk fungsi \( f(x) = \frac{1}{2}x^{2} - 3x \): Menggunakan prinsip yang sama, kita peroleh: \[ f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] Dengan \( f(x+h) = \frac{1}{2}(x+h)^{2} - 3(x+h) \): \[ = \frac{1}{2}(x^{2} + 2xh + h^{2}) - 3(x + h) = \frac{1}{2}x^{2} + xh + \frac{1}{2}h^{2} - 3x - 3h \] Jadi, \[ f^{\prime}(x) = \lim_{h \to 0} \frac{\left(\frac{1}{2}x^{2} + xh + \frac{1}{2}h^{2} - 3x - 3h\right) - \left(\frac{1}{2}x^{2} - 3x\right)}{h} = \lim_{h \to 0} \frac{xh + \frac{1}{2}h^{2} - 3h}{h} = \lim_{h \to 0} (x - 3 + \frac{1}{2}h) = x - 3 \] Maka, terbitan pertama bagi fungsi ini ialah \( f^{\prime}(x) = x - 3 \).