Tugas 3 Kalkulus: Diferensial (Turunan) Fungsi Trigonometri

a. \(y = 6 \sin 5x + 4 \cos 4x\)

\[\begin{aligned} y' &= \frac{d}{dx}(6 \sin 5x) + \frac{d}{dx}(4 \cos 4x) \\ &= 6(\cos 5x \cdot 5) + 4(-\sin 4x \cdot 4) \\ &= 30 \cos 5x - 16 \sin 4x \end{aligned}\]

b. \(y = 4 \sin^4 x + 3 \cos^3 x\)

\[\begin{aligned} y' &= 4 \cdot 4\sin^{4-1}x \cdot \frac{d}{dx}(\sin x) + 3 \cdot 3\cos^{3-1}x \cdot \frac{d}{dx}(\cos x) \\ &= 16 \sin^3 x \cos x + 9 \cos^2 x (-\sin x) \\ &= 16 \sin^3 x \cos x - 9 \cos^2 x \sin x \end{aligned}\]

c. \(y = 5 \sin^3 x \cos^5 x\)

Gunakan Aturan Perkalian: \((u \cdot v)' = u'v + uv'\) \[\begin{aligned} u &= 5 \sin^3 x \implies u' = 15 \sin^2 x \cos x \\ v &= \cos^5 x \implies v' = 5 \cos^4 x (-\sin x) = -5 \cos^4 x \sin x \\ y' &= (15 \sin^2 x \cos x)(\cos^5 x) + (5 \sin^3 x)(-5 \cos^4 x \sin x) \\ &= 15 \sin^2 x \cos^6 x - 25 \sin^4 x \cos^4 x \\ &= 5 \sin^2 x \cos^4 x (3 \cos^2 x - 5 \sin^2 x) \quad \text{(Bentuk Sederhana)} \end{aligned}\]

d. \(y = \frac{3x^3 \sin^4 x}{2 \cos^3 x}\)

Gunakan Aturan Pembagian: \(y' = \frac{u'v - uv'}{v^2}\) \[\begin{aligned} u &= 3x^3 \sin^4 x \implies u' = 9x^2 \sin^4 x + 12x^3 \sin^3 x \cos x \\ v &= 2 \cos^3 x \implies v' = -6 \cos^2 x \sin x \\ y' &= \frac{(9x^2 \sin^4 x + 12x^3 \sin^3 x \cos x)(2 \cos^3 x) - (3x^3 \sin^4 x)(-6 \cos^2 x \sin x)}{(2 \cos^3 x)^2} \\ &= \frac{18x^2 \sin^4 x \cos^3 x + 24x^3 \sin^3 x \cos^4 x + 18x^3 \sin^5 x \cos^2 x}{4 \cos^6 x} \end{aligned}\]

Bentuk sederhana (faktorkan \(6x^2 \sin^3 x \cos^2 x\)): \[\begin{aligned} y' = \frac{3x^2 \sin^3 x (3 \sin x \cos x + 4x \cos^2 x + 3x \sin^2 x)}{2 \cos^4 x} \end{aligned}\]

e. \(y = \frac{5 \cos 2x + 3 \sin 4x}{3 \cos 5x + \sin 6x}\)

Gunakan Aturan Pembagian. \[\begin{aligned} u &= 5 \cos 2x + 3 \sin 4x \implies u' = -10 \sin 2x + 12 \cos 4x \\ v &= 3 \cos 5x + \sin 6x \implies v' = -15 \sin 5x + 6 \cos 6x \\ y' &= \frac{(-10 \sin 2x + 12 \cos 4x)(3 \cos 5x + \sin 6x) - (5 \cos 2x + 3 \sin 4x)(-15 \sin 5x + 6 \cos 6x)}{(3 \cos 5x + \sin 6x)^2} \end{aligned}\]

(Pembilang tidak disederhanakan lebih lanjut karena akan menjadi sangat panjang dan tidak memiliki faktor persekutuan yang jelas).

f. \(y = \frac{2x^3 + x^2}{x^2 \sin 5x}\)

Gunakan Aturan Pembagian. \[\begin{aligned} u &= 2x^3 + x^2 \implies u' = 6x^2 + 2x \\ v &= x^2 \sin 5x \implies v' = 2x \sin 5x + 5x^2 \cos 5x \\ y' &= \frac{(6x^2 + 2x)(x^2 \sin 5x) - (2x^3 + x^2)(2x \sin 5x + 5x^2 \cos 5x)}{(x^2 \sin 5x)^2} \\ &= \frac{(6x^4 + 2x^3)\sin 5x - (4x^4 + 2x^3)\sin 5x - (10x^5 + 5x^4)\cos 5x}{x^4 \sin^2 5x} \\ &= \frac{2x^4 \sin 5x - (10x^5 + 5x^4)\cos 5x}{x^4 \sin^2 5x} \\ &= \frac{2 \sin 5x - 5(2x+1)\cos 5x}{\sin^2 5x} \quad \text{(Sederhana)} \end{aligned}\]

g. \(y = 5 \cos^3 4x + \sin^2 2x\)

\[\begin{aligned} y' &= 5 \cdot 3 \cos^2 4x \cdot (-\sin 4x) \cdot 4 + 2 \sin 2x \cdot (\cos 2x) \cdot 2 \\ &= -60 \cos^2 4x \sin 4x + 4 \sin 2x \cos 2x \\ &= -60 \cos^2 4x \sin 4x + 2 \sin 4x \quad \text{(Karena } 2\sin 2x\cos 2x = \sin 4x) \end{aligned}\]

h. \(y = 4 \cos^5 (x^2 + 3x)\)

Gunakan Aturan Rantai (Turunan luar, turunan trigonometri, turunan dalam). \[\begin{aligned} y' &= 4 \cdot 5 \cos^4 (x^2 + 3x) \cdot (-\sin (x^2 + 3x)) \cdot (2x + 3) \\ &= -20(2x + 3) \cos^4 (x^2 + 3x) \sin (x^2 + 3x) \end{aligned}\]

i. \(y = \sqrt[3]{\sin^2 3x}\)

Ubah ke bentuk pangkat: \(y = (\sin 3x)^{2/3}\) \[\begin{aligned} y' &= \frac{2}{3} (\sin 3x)^{\frac{2}{3} - 1} \cdot \frac{d}{dx}(\sin 3x) \\ &= \frac{2}{3} (\sin 3x)^{-1/3} \cdot (\cos 3x \cdot 3) \\ &= 2 (\sin 3x)^{-1/3} \cos 3x \\ &= \frac{2 \cos 3x}{\sqrt[3]{\sin 3x}} \end{aligned}\]

j. \(y = 2x^2 \sin^3 x - 4x^3 \cos 5x\)

Gunakan Aturan Perkalian untuk masing-masing suku. \[\begin{aligned} y' &= \left[ (4x)(\sin^3 x) + (2x^2)(3 \sin^2 x \cos x) \right] - \left[ (12x^2)(\cos 5x) + (4x^3)(-5 \sin 5x) \right] \\ &= (4x \sin^3 x + 6x^2 \sin^2 x \cos x) - (12x^2 \cos 5x - 20x^3 \sin 5x) \\ &= 4x \sin^3 x + 6x^2 \sin^2 x \cos x - 12x^2 \cos 5x + 20x^3 \sin 5x \end{aligned}\]