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Diketahui fungsi \( f(x)=\sin 2 x+\cos 3 x \). Jika \( f^{\prime}(x) \) adalah turunan pertama dari \( f(x) \), hitunglah nilai \( f^{\prime}\left(\frac{\pi}{6}\right) \). A. -3 B. -2 C. 0 D. 1 E. 2

Ask by Grant Parry. in Indonesia
Feb 25,2025

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Tutor-Verified Answer

Answer

\( f^{\prime}\left(\frac{\pi}{6}\right) = -2 \), so the answer is B.

Solution

We are given the function f(x) = sin(2x) + cos(3x). Its first derivative f '(x) is found by differentiating each term:   f '(x) = 2 cos(2x) - 3 sin(3x). To find f '(π/6), substitute x = π/6:   f '(π/6) = 2 cos(2·(π/6)) - 3 sin(3·(π/6))        = 2 cos(π/3) - 3 sin(π/2). Recall the trigonometric values:   cos(π/3) = 1/2  and sin(π/2) = 1. Thus:   f '(π/6) = 2 · (1/2) - 3 · 1        = 1 - 3        = -2. Therefore, the value of f '(π/6) is -2. The correct answer is Option B.

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The Deep Dive

Untuk menemukan \( f^{\prime}(x) \), kita terlebih dahulu perlu menghitung turunan dari fungsi \( f(x) = \sin 2x + \cos 3x \). Menggunakan aturan turunan, kita dapat menghitung: \[ f^{\prime}(x) = \frac{d}{dx}(\sin 2x) + \frac{d}{dx}(\cos 3x) \] \[ = 2 \cos 2x - 3 \sin 3x \] Selanjutnya, kita substitusikan \( x = \frac{\pi}{6} \) ke dalam turunan \( f^{\prime}(x) \): \[ f^{\prime}\left(\frac{\pi}{6}\right) = 2 \cos\left(2 \cdot \frac{\pi}{6}\right) - 3 \sin\left(3 \cdot \frac{\pi}{6}\right) \] \[ = 2 \cos\left(\frac{\pi}{3}\right) - 3 \sin\left(\frac{\pi}{2}\right) \] Dengan nilai \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \) dan \( \sin\left(\frac{\pi}{2}\right) = 1 \), kita mendapatkan: \[ f^{\prime}\left(\frac{\pi}{6}\right) = 2 \cdot \frac{1}{2} - 3 \cdot 1 = 1 - 3 = -2 \] Jadi, nilai \( f^{\prime}\left(\frac{\pi}{6}\right) \) adalah: **B. -2**

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