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find the deri...

Question

find the derivative of sin^3(3x+5) using first principles

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Solution

$f (x) = \sin^{3} (3 x + 5) \frac{d y}{d x} = \lim_{∆ x \to 0} \frac{f (x + ∆ x) - f (x)}{∆ x} = \lim_{∆ x \to 0} \frac{\sin^{3} (3 (x + ∆ x) + 5) - \sin^{3} (3 x + 5)}{∆ x} [S i n c e a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)] = \lim_{∆ x \to 0} \frac{\{\sin (3 (x + ∆ x) + 5) - \sin (3 x + 5)\} \{\sin^{2} (3 (x + ∆ x) + 5) + \sin^{2} (3 x + 5) + \sin (3 (x + ∆ x) + 5) \sin (3 x + 5)\}}{∆ x} [S i n C - \sin D = 2 \sin \frac{C - D}{2} \cos \frac{C + D}{2}] = \lim_{∆ x \to 0} \frac{\{2 \sin \frac{(3 (x + ∆ x) + 5 - (3 x + 5))}{2} \cos \frac{(3 (x + ∆ x) + 5 + (3 x + 5))}{2}\} \{\sin^{2} (3 (x + ∆ x) + 5) + \sin^{2} (3 x + 5) + \sin (3 (x + ∆ x) + 5) \sin (3 x + 5)\}}{∆ x} = \lim_{∆ x \to 0} \frac{\{2 \sin \frac{3 \times ∆ x}{2} \cos \frac{(6 x + 3 ∆ x + 10)}{2}\} \{\sin^{2} (3 (x + ∆ x) + 5) + \sin^{2} (3 x + 5) + \sin (3 (x + ∆ x) + 5) \sin (3 x + 5)\}}{∆ x} = \lim_{∆ x \to 0} [\frac{\{2 \sin \frac{3 \times ∆ x}{2} \cos \frac{(6 x + 3 ∆ x + 10)}{2}\}}{∆ x} \{\sin^{2} (3 (x + ∆ x) + 5) + \sin^{2} (3 x + 5) + \sin (3 (x + ∆ x) + 5) \sin (3 x + 5)\}] = \lim_{∆ x \to 0} \frac{2 \sin \frac{3 \times ∆ x}{2}}{∆ x} \lim_{∆ x \to 0} \cos \frac{(6 x + 3 ∆ x + 10)}{2} \lim_{∆ x \to 0} \{\sin^{2} (3 (x + ∆ x) + 5) + \sin^{2} (3 x + 5) + \sin (3 (x + ∆ x) + 5) \sin (3 x + 5)\} = \lim_{∆ x \to 0} \frac{\sin \frac{3 \times ∆ x}{2}}{\frac{∆ x}{2}} \lim_{∆ x \to 0} \cos \frac{(6 x + 3 ∆ x + 10)}{2} \lim_{∆ x \to 0} \{\sin^{2} (3 (x + ∆ x) + 5) + \sin^{2} (3 x + 5) + \sin (3 (x + ∆ x) + 5) \sin (3 x + 5)\} = \lim_{∆ x \to 0} \frac{3 \sin \frac{3 \times ∆ x}{2}}{\frac{3 \times ∆ x}{2}} \lim_{∆ x \to 0} \cos \frac{(6 x + 3 ∆ x + 10)}{2} \lim_{∆ x \to 0} \{\sin^{2} (3 (x + ∆ x) + 5) + \sin^{2} (3 x + 5) + \sin (3 (x + ∆ x) + 5) \sin (3 x + 5)\} = 3 \lim_{∆ x \to 0} \frac{\sin \frac{3 \times ∆ x}{2}}{\frac{3 \times ∆ x}{2}} \lim_{∆ x \to 0} \cos \frac{(6 x + 3 ∆ x + 10)}{2} \lim_{∆ x \to 0} \{\sin^{2} (3 (x + ∆ x) + 5) + \sin^{2} (3 x + 5) + \sin (3 (x + ∆ x) + 5) \sin (3 x + 5)\} [W e k n o w \lim_{θ \to 0} \frac{\sin θ}{θ} = 1] = 3 \times 1 \times \cos \frac{(6 x + 10)}{2} \{\sin^{2} (3 x + 5) + \sin^{2} (3 x + 5) + \sin (3 x + 5) \sin (3 x + 5)\} = 3 \times 1 \times \cos (3 x + 5) \{\sin^{2} (3 x + 5) + \sin^{2} (3 x + 5) + \sin^{2} (3 x + 5)\} = 3 \times 1 \times \cos (3 x + 5) \{3 \sin^{2} (3 x + 5)\} = 9 \sin^{2} (3 x + 5) \cos (3 x + 5)$

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