Limits and Derivatives: Derivative of Trigonometric Function from First Principles

$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{\sqrt{\sin(x + h)} - \sqrt{\sin x}}{h}$$

Multiply numerator and denominator by $\sqrt{\sin(x + h)} + \sqrt{\sin x}$: $$f'(x) = \lim_{h \to 0} \frac{[\sin(x + h) - \sin x]}{h [\sqrt{\sin(x + h)} + \sqrt{\sin x}]}$$

Using $\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)$: $$\sin(x + h) - \sin x = 2 \cos\left(x + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)$$ $$f'(x) = \lim_{h \to 0} \frac{2 \cos(x + h/2) \sin(h/2)}{h [\sqrt{\sin(x + h)} + \sqrt{\sin x}]}$$ $$= \lim_{h \to 0} \cos\left(x + \frac{h}{2}\right) \times \lim_{h \to 0} \left[ \frac{\sin(h/2)}{h/2} \right] \times \lim_{h \to 0} \frac{1}{\sqrt{\sin(x + h)} + \sqrt{\sin x}}$$

- $\lim_{h \to 0} \cos(x + h/2) = \cos x$ - $\lim_{h \to 0} \frac{\sin(h/2)}{h/2} = 1$ - $\lim_{h \to 0} \frac{1}{\sqrt{\sin(x + h)} + \sqrt{\sin x}} = \frac{1}{2\sqrt{\sin x}}$ $$f'(x) = \frac{\cos x}{2\sqrt{\sin x}}$$