1. Muammo: Agar $f^3(2x + 1) = x^3 + 3x + 4$ bo'lsa, $f'(3)$ ning qiymatini toping. 2. Bu yerda $f^3$ deganda $f$ funksiyaning kubi tushuniladi, ya'ni $f^3(t) = (f(t))^3$. 3. Berilgan tenglama: $$(f(2x + 1))^3 = x^3 + 3x + 4$$ 4. Ikkala tomonning hosilasini $x$ bo'yicha olamiz. Chap tomonda zanjir qoidasi bo'yicha: $$3(f(2x + 1))^2 \cdot f'(2x + 1) \cdot 2 = 3(f(2x + 1))^2 \cdot f'(2x + 1) \cdot 2$$ 5. O'ng tomonda: $$3x^2 + 3$$ 6. Hosilani tenglashtiramiz: $$3(f(2x + 1))^2 \cdot f'(2x + 1) \cdot 2 = 3x^2 + 3$$ 7. Soddalashtiramiz: $$6(f(2x + 1))^2 f'(2x + 1) = 3x^2 + 3$$ 8. $f'(2x + 1)$ ni ifodalash uchun: $$f'(2x + 1) = \frac{3x^2 + 3}{6(f(2x + 1))^2} = \frac{x^2 + 1}{2(f(2x + 1))^2}$$ 9. $f'(3)$ ni topish uchun $2x + 1 = 3$ bo'lishi kerak, ya'ni: $$2x + 1 = 3 \Rightarrow 2x = 2 \Rightarrow x = 1$$ 10. $f(3)$ ni topamiz. Berilgan tenglamadan $x=1$ ni qo'yamiz: $$(f(3))^3 = 1^3 + 3 \cdot 1 + 4 = 1 + 3 + 4 = 8$$ 11. Demak: $$f(3) = \sqrt[3]{8} = 2$$ 12. Endi $f'(3)$ ni hisoblaymiz: $$f'(3) = \frac{1^2 + 1}{2 \cdot (2)^2} = \frac{2}{2 \cdot 4} = \frac{2}{8} = \frac{1}{4}$$ Javob: $f'(3) = \frac{1}{4}$