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derivative using first principle -

sin_/x (sin under root x)

$Let f (x) = sin \sqrt{x} Then, f (x + h) = \sin \sqrt{x + h} ∴ \frac{d}{d x} [f (x)] = \lim_{h \to 0} \frac{f (x + h) - f (h)}{h} \Rightarrow \frac{d}{d x} [f (x)] = \lim_{h \to 0} \frac{\sin \sqrt{x + h} - \sin \sqrt{x}}{h} \frac{d}{d x} [f (x)] = \lim_{h \to 0} \frac{\frac{2 \sin (\sqrt{x + h} - \sqrt{x})}{2} \cos (\frac{\sqrt{x + h} + \sqrt{x}}{2})}{h} = \lim_{h \to 0} \sin (\frac{\sqrt{x + h} - \sqrt{x}}{2}) \cdot \frac{(\sqrt{x + h} - \sqrt{x}) (\sqrt{x + h} + \sqrt{x})}{(\sqrt{x + h} + \sqrt{x}) h} \times \cos (\frac{\sqrt{x + h} + \sqrt{x}}{2}) = \lim_{h \to 0} \frac{\sin (\frac{\sqrt{x + h} - \sqrt{x}}{2})}{(\frac{\sqrt{x + h} - \sqrt{x}}{2})} \lim_{h \to 0} \frac{x + h - x}{(\sqrt{x + h} + \sqrt{x}) h} \lim_{h \to 0} \cos (\frac{\sqrt{x + h} + \sqrt{x}}{2}) = \lim_{θ \to 0} \frac{\sin θ}{θ} \cdot \lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}} \lim_{h \to 0} \cos (\frac{\sqrt{x + h} + \sqrt{x}}{2}) where θ = \frac{\sqrt{x + h} - \sqrt{x}}{2} \Rightarrow \frac{d}{d x} [f (x)] = 1 \times \frac{1}{2 \sqrt{x}} \times \cos (\sqrt{x}) ∴ \frac{d}{d x} (\sin \sqrt{x}) = \frac{\cos \sqrt{x}}{2 \sqrt{x}}$

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