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

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