Differentiate √cosecx from the first principle

$\begin{aligned} & \text { Let } y=f(x)=\operatorname{cosec} x \\ & \therefore f(x+h)=\operatorname{cosec}(x+h) \\ & \therefore \frac{d y}{d t}=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \rightarrow 0} \frac{\operatorname{cosec}(x+h)-\operatorname{cosec} x}{h} \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{1}{\sin (x+h)}-\frac{1}{\sin x}\right] \\ & =\lim _{h \rightarrow 0} \frac{\sin x-\sin (x+h)}{h \cdot \sin x \cdot \sin (x+h)} \\ & =\lim _{h \rightarrow 0} \frac{2 \cos \left(\frac{2 x+h}{2}\right) \sin \left(-\frac{h}{2}\right)}{h \cdot \sin x \sin (x+h)} \\ & =-\lim _{h \rightarrow 0} \frac{\cos \left(x+\frac{h}{h}\right)}{\sin x \cdot \sin (u+h)} \cdot \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{h / 2} \\ & =-\frac{1 \cos x}{\sin x \cdot \sin x} \cdot \lim _{z \rightarrow 0} \frac{\sin z}{z} \\ & {\left[z=\frac{h}{2} \text {; Then, } z \rightarrow 0 \text { when } \Rightarrow h \rightarrow 0\right]} \\ & =\frac{-\cos x}{\sin x \cdot \sin x} \cdot 1 \\ & =-\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} \\ & =-\operatorname{cosec} x . \cot x \\ & \therefore \frac{d y}{d x}=-1 \operatorname{csec} x \cdot \cot x \\ & \end{aligned}$