Differentiate sinx/x from the first principle - Maths - Limits and Derivatives

Question

Differentiate sinx/x from the first principle

Ankush Jain · Accepted Answer

Let $f (x) = \frac{\sin x}{x}$ Then, $f (x + h) = \frac{\sin (x + h)}{x + h}$ $∴ \frac{d}{d x} f (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \Rightarrow \frac{d}{d x} f (x) = \lim_{h \to 0} \frac{\frac{\sin (x + h)}{x + h} - \frac{\sin x}{x}}{h} = \lim_{h \to 0} \frac{x \sin (x + h) - (x + h) \sin x}{h x (x + h)} = \lim_{h \to 0} \frac{x \sin (x + h) - x \sin x - h \sin x}{h x (x + h)} = \lim_{h \to 0} \frac{x [\sin (x + h) - \sin x] - h \sin x}{h x (x + h)} = \lim_{h \to 0} \frac{x [2 \cos \frac{2 x + h}{2} \sin \frac{h}{2}] - h \sin x}{h x (x + h)} [u \sin g : \sin C - \sin D = 2 \cos \frac{C + D}{2} \sin \frac{C - D}{2}] = \lim_{h \to 0} \frac{x [2 \cos \frac{2 x + h}{2} \sin \frac{h}{2}]}{h x (x + h)} - \lim_{h \to 0} \frac{h \sin x}{h x (x + h)} = \lim_{h \to 0} \frac{[2 \cos \frac{2 x + h}{2} \sin \frac{h}{2}]}{h (x + h)} - \lim_{h \to 0} \frac{\sin x}{x (x + h)} = \lim_{h \to 0} \frac{[\cos (x + \frac{h}{2}) \sin \frac{h}{2}]}{(x + h) \frac{h}{2}} - \lim_{h \to 0} \frac{\sin x}{x (x + h)} = \lim_{h \to 0} \frac{[\cos (x + \frac{h}{2})]}{(x + h)} \lim_{h \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} - \lim_{h \to 0} \frac{\sin x}{x (x + h)} = \frac{\cos (x + 0)}{x} 1 - \frac{\sin x}{x x} = \frac{\cos x}{x} - \frac{\sin x}{x^{2}} = \frac{x \cos x - \sin x}{x^{2}}$

Differentiate sinx/x from the first principle.