Soal 01
Jika \(f(x) = 2x^2 + ax \:-\:b\) dan \(f(2) = 8\), \(f'(1) = 5\), maka nilai \(a + b = \dotso\)
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Answer: C
\(f(2) = 8\)
\(2(2)^2 + a(2)\:-\:b = 8\)
\(8 + 2a\:-\:b = 8\)
\(2a\:-\:b = 0 \dotso \color{red} (1)\)
\(f'(x) = 4x + a\)
\(f'(1) = 5\)
\(4(1) + a = 5\)
\(a = 1 \dotso \color{red} (2)\)
Substitusikan persamaan (2) ke persamaan (1)
\(2(1)\:-\:b = 0\)
\(b = 2\)
\(a + b = 1 + 2 = 3\)
Soal 02
Diketahui \(f(x) = \dfrac{x^2 + 4}{4x + 1}\). Jika \(f'(x)\) menyatakan turunan pertama \(f(x)\), maka \(f(0) + f'(0) = \dotso\)
(A) \(-12\)
(B) \(-11\)
(C) \(-10\)
(D) \(9\)
(E) \(12\)
Answer: A
\(f(0) = \dfrac{0^2 + 4}{4(0) + 1} = 4\)
Menentukan \(f'(x)\).
\(f(x) = \dfrac{x^2 + 4}{4x + 1}\)
Misal :
\(u = x^2 + 4 \rightarrow u’ = 2x\)
\(v = 4x + 1 \rightarrow v’ = 4\)
\(\color{blue} f'(x) = \dfrac{u’ \cdot v \:-\: u \cdot v’}{v^2}\)
\(f'(x) = \dfrac{2x\cdot (4x + 1) \:-\: (x^2 + 4) \cdot 4}{(4x + 1)^2}\)
\(f'(0) = \dfrac{2(0)\cdot (4(0) + 1) \:-\: ((0)^2 + 4) \cdot 4}{(4(0) + 1)^2}\)
\(f'(0) = \dfrac{0 \:-\: 16}{1} = -16\)
\(f(0) + f'(0) = 4 + (-16) = -12\)
Soal 03
Bila \(f(x) = \dfrac{(x^2 \:-\:4x + 2)^3}{2x + 1}\) maka \(f'(x) = \dotso\)
(A) \(\dfrac{2(x^2 + 4x + 2)^2(5x^2\:-\:5x\:-\:8)}{(2x + 1)^2}\)
(B) \(\dfrac{2(x^2\:-\:4x + 2)^2(5x^2\:-\:5x\:-\:8)}{(2x + 1)^2}\)
(C) \(\dfrac{4(x^2\:-\:4x + 2)^2(5x^2\:-\:5x\:-\:8)}{(2x + 1)^2}\)
(D) \(\dfrac{2(x^2\:-\:4x + 2)^2(5x^2\:-\:5x\:-\:8)}{(2x \:-\: 1)^2}\)
(E) \(\dfrac{2(x^2\:-\:4x + 2)(5x^2\:-\:5x\:-\:8)}{(2x + 1)^3}\)
Answer: B
Misal:
\(u = (x^2 \:-\:4x + 2)^3 \rightarrow u’ = 6(x\:-\:2)(x^2\:-\:4x + 2)^2\)
\(v = 2x + 1 \rightarrow v’ = 2\)
\(\color{blue} f'(x) = \dfrac{u’ \cdot v \:-\: u \cdot v’}{v^2}\)
\(f'(x) = \dfrac{6(x\:-\:2)(x^2\:-\:4x + 2)^2 \cdot (2x + 1)\:-\:(x^2 \:-\:4x + 2)^3 \cdot 2}{(2x + 1)^2}\)
\(f'(x) = \dfrac{2(x^2\:-\:4x + 2)^2(3(x\:-\:2)(2x + 1)\:-\:(x^2 \:-\:4x + 2))}{(2x + 1)^2}\)
\(f'(x) = \dfrac{2(x^2\:-\:4x + 2)^2(3(2x^2\:-\:3x\:-\:2)\:-\:x^2 + 4x \:-\:2)}{(2x + 1)^2}\)
\(f'(x) = \dfrac{2(x^2\:-\:4x + 2)^2(6x^2\:-\:9x\:-\:6\:-\:x^2 + 4x \:-\:2)}{(2x + 1)^2}\)
\(f'(x) = \dfrac{2(x^2\:-\:4x + 2)^2(5x^2\:-\:5x\:-\:8)}{(2x + 1)^2}\)
Soal 04
Turunan pertama dari \(y = \sqrt[3] {6x^2 + 9x}\) adalah…
(A) \(\dfrac{x + 3}{\sqrt[3] {(3x^2 + 2x)^2}}\)
(B) \(\dfrac{2x + 3}{\sqrt[3] {(3x^2 + 9x)^2}}\)
(C) \(\dfrac{4x + 1}{\sqrt[3] {(6x^2 + 9x)^2}}\)
(D) \(\dfrac{4x \:-\: 3}{\sqrt[3] {(6x^2 + 9x)^2}}\)
(E) \(\dfrac{4x + 3}{\sqrt[3] {(6x^2 + 9x)^2}}\)
Answer: E
\(y = (6x^2 + 9x)^{\dfrac{1}{3}}\)
\(y’ = (12x +9)\cdot \dfrac{1}{3} \cdot (6x^2 + 9x)^{-\dfrac{2}{3}}\)
\(y’ = \dfrac{4x + 3}{\sqrt[3] {(6x^2 + 9x)^2}}\)
Soal 05
Jika \(y = 4u^5\), \(u = 2v\:-\:1\), dan \(v = \sqrt[4]{x} + 5\), maka \(\dfrac{dy}{dx} = \dotso\)
(A) \(\dfrac{5(2\sqrt[4] {x} + 9)^3}{\sqrt[4] {x^3}}\)
(B) \(\dfrac{10(\sqrt[4] {x} \:-\: 9)^4}{\sqrt[4] {x^3}}\)
(C) \(\dfrac{10(2\sqrt[4] {x} + 9)^4}{\sqrt[4] {x^3}}\)
(D) \(\dfrac{12(2\sqrt[4] {x} + 9)^4}{\sqrt[4] {x^3}}\)
(E) \(\dfrac{15(2\sqrt[4] {x} + 9)^4}{\sqrt[4] {x^3}}\)
Answer: C
\(y = 4u^5 \rightarrow \dfrac{dy}{du} = 20u^4\)
\(u = 2v\:-\:1 \rightarrow \dfrac{du}{dv} = 2\)
\(v = \sqrt[4]{x} + 5 \rightarrow \dfrac{dv}{dx} = \dfrac{1}{4 \sqrt[4] {x^3}}\)
\(\color{blue} \dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dv} \times \dfrac{dv}{dx}\)
\(\dfrac{dy}{dx} = 20u^4 \cdot 2 \cdot \dfrac{1}{4 \sqrt[4] {x^3}}\)
\(\dfrac{dy}{dx} = \dfrac{10u^4}{\sqrt[4] {x^3}}\)
\(\dfrac{dy}{dx} = \dfrac{10(2\sqrt[4] {x} + 9)^4}{\sqrt[4] {x^3}}\)
Soal 06
Jika \(f(x) = \dfrac{6}{x^3}\), maka \(\lim\limits_{h \rightarrow 0} \dfrac{f(3x + 2h)\:-\:f(3x)}{h} = \dotso\)
(A) \(-\dfrac{4}{9x^4}\)
(B) \(-\dfrac{8}{9x^4}\)
(C) \(-\dfrac{16}{9x^4}\)
(D) \(-\dfrac{4}{3x^4}\)
(E) \(-\dfrac{8}{3x^4}\)
Answer: A
\(\color{blue} \lim\limits_{h \rightarrow 0} \dfrac{f(ax + bh)\:-\:f(ax)}{h} = b\cdot f'(ax)\)
\(f(3x) = \dfrac{6}{(3x)^3}\)
\(f(3x) = \dfrac{6}{27x^3}\)
\(f(3x) = \dfrac{2}{9x^3}\)
\(f(3x) = \dfrac{2}{9}x^{-3}\)
\(3 \cdot f'(3x) = \dfrac{2}{9} \cdot (-3) \cdot x^{-4}\)
\(f'(3x) = -\dfrac{2}{9} x^{-4}\)
\(\lim\limits_{h \rightarrow 0} \dfrac{f(3x + 2h)\:-\:f(3x)}{h} = 2 \cdot f'(3x)\)
\(\lim\limits_{h \rightarrow 0} \dfrac{f(3x + 2h)\:-\:f(3x)}{h} = 2 \cdot \left( -\dfrac{2}{9} x^{-4}\right)\)
\(\lim\limits_{h \rightarrow 0} \dfrac{f(3x + 2h)\:-\:f(3x)}{h} = -\dfrac{4}{9x^4}\)
Soal 07
Turunan pertama dari \(f(x) = (5x\:-\:2)^3(3x\:-\:1)^5\) adalah…
(A) \(f'(x) = 10(8x\:-\:3)(5x\:-\:2)^2 (3x\:-\:1)^4\)
(B) \(f'(x) = 12(8x\:-\:3)(5x\:-\:2)^2 (3x\:-\:1)^4\)
(C) \(f'(x) = 15(8x\:-\:3)(5x\:-\:2)^2 (3x\:-\:1)^4\)
(D) \(f'(x) = 16(8x\:-\:3)(5x\:-\:2)^2 (3x\:-\:1)^4\)
(E) \(f'(x) = 18(8x\:-\:3)(5x\:-\:2)^2 (3x\:-\:1)^4\)
‘
Answer: C
Misal:
\(u = (5x\:-\:2)^3 \rightarrow u’ = 15(5x\:-\:2)^2\)
\(v = (3x\:-\:1)^5 \rightarrow v’ = 15(3x\:-\:1)^4\)
\(\color{blue} f'(x) = u’ \cdot v + u \cdot v’\)
\(f'(x) = 15(5x\:-\:2)^2 \cdot (3x\:-\:1)^5 + (5x\:-\:2)^3\cdot 15(3x\:-\:1)^4\)
Selanjutnya faktorkan,
\(f'(x) = 15(5x\:-\:2)^2 (3x\:-\:1)^4(3x\:-\:1 + 5x\:-\:2)\)
\(f'(x) = 15(8x\:-\:3)(5x\:-\:2)^2 (3x\:-\:1)^4\)
Soal 08
Jika \(f(x) = \sqrt{\dfrac{2x\:-\:5}{3x + 1}}\), maka \(f'(x) = \dotso\)
(A) \(\dfrac{15}{2(3x+1)\sqrt{(2x\:-\:5)(3x + 1)}}\)
(B) \(\dfrac{16}{2(3x+1)\sqrt{(2x\:-\:5)(3x + 1)}}\)
(C) \(\dfrac{17}{2(3x+1)\sqrt{(2x\:-\:5)(3x + 1)}}\)
(D) \(\dfrac{18}{2(3x+1)\sqrt{(2x\:-\:5)(3x + 1)}}\)
(E) \(\dfrac{19}{2(3x+1)\sqrt{(2x\:-\:5)(3x + 1)}}\)
Answer: C
\(f(x) = \left( \dfrac{2x\:-\:5}{3x + 1}\right)^{\dfrac{1}{2}}\)
\(f'(x) = \dfrac{2(3x+1)\:-\:(2x\:-\:5)3}{(3x + 1)^2}\cdot \dfrac{1}{2} \cdot \left(\dfrac{2x\:-\:5}{3x + 1} \right)^{-\dfrac{1}{2}}\)
\(f'(x) = \dfrac{17}{2(3x + 1)^2}\sqrt{\dfrac{3x + 1}{2x\:-\:5}}\)
\(f'(x) = \dfrac{17}{2(3x+1)\sqrt{(2x\:-\:5)(3x + 1)}}\)
Soal 09
Jika \(f(x) = \sqrt{(8x + 2)(4x^2\:-\:1)}\), maka \(f'(x) = \dotso\)
(A) \(\dfrac{48x^2 + 4x\:-\:4}{\sqrt{(8x + 2)(4x^2\:-\:1)}}\)
(B) \(\dfrac{43x^2 + 8x\:-\:4}{\sqrt{(8x + 2)(4x^2\:-\:1)}}\)
(C) \(\dfrac{48x^2 \:-\: 8x\:-\:4}{\sqrt{(8x + 2)(4x^2\:-\:1)}}\)
(D) \(\dfrac{48x^2 + 8x\:-\:4}{\sqrt{(8x + 2)(4x^2\:-\:1)}}\)
(E) \(\dfrac{49x^2 + 8x\:-\:4}{\sqrt{(8x + 2)(4x^2\:-\:1)}}\)
Answer: D
\(f(x) = \left[(8x + 2)(4x^2\:-\:1)\right]^{\dfrac{1}{2}}\)
\(f'(x) = [8(4x^2\:-\:1) + (8x + 2)(8x)]\cdot \dfrac{1}{2} \cdot [(8x + 2)(4x^2\:-\:1)]^{-\dfrac{1}{2}}\)
\(f'(x) = \dfrac{32x^2\:-\:8 + 64x^2 + 16x}{2\sqrt{(8x + 2)(4x^2\:-\:1)}}\)
\(f'(x) = \dfrac{96x^2 + 16x \:-\:8}{2\sqrt{(8x + 2)(4x^2\:-\:1)}}\)
\(f'(x) = \dfrac{48x^2 + 8x \:-\:4}{\sqrt{(8x + 2)(4x^2\:-\:1)}}\)
Soal 10
Diketahui \(f(x) = \dfrac{4x + 3}{6x\:-\:2}\), jika \(f'(x) = \dfrac{-A}{(Bx\:-\:2)^2}\) maka \(A\:-\:B = \dotso\)
(A) 10
(B) 12
(C) 15
(D) 18
(E) 20
Answer: E
\(f'(x) = \dfrac{4(6x\:-\:2)\:-\:(4x + 3)6}{(6x\:-\:2)^2}\)
\(f'(x) = \dfrac{24x\:-\:8\:-\:24x\:-\:18}{(6x\:-\:2)^2}\)
\(f'(x) = \dfrac{-26}{(6x\:-\:2)^2}\)
\(A = 26\)
\(B = 6\)
\(A\:-\:B = 26\:-\:6 = 20\)
