Question

how to find the derivative of cosec x by the first principle method ?/

Answer 1

$$\begin{gathered} y=\cos ecx \\ \frac{dy}{dx}=\underset{h\to 0}{\lim }\frac{\cos ec\left(x+h\right)-\cos ecx}{h} \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{\sin \left(x+h\right)}}-{\displaystyle \frac{1}{\sin x}}}{h} \\ =\underset{h\to 0}{\lim }\frac{\sin x-\sin \left(x+h\right)}{h\sin x\sin \left(x+h\right)} \\ =\underset{h\to 0}{\lim }\frac{2\cos \left({\displaystyle \frac{x+x+h}{2}}\right)\sin \left({\displaystyle \frac{x-x-h}{2}}\right)}{h\sin x\sin \left(x+h\right)} \\ =\underset{h\to 0}{\lim }\frac{-2\cos \left({\displaystyle x+\frac{h}{2}}\right)\sin \left({\displaystyle \frac{h}{2}}\right)}{h\sin x\sin \left(x+h\right)} \\ =\underset{h\to 0}{\lim }\frac{-\cos \left({\displaystyle x+\frac{h}{2}}\right)}{\sin x\sin \left(x+h\right)}\times \underset{h\to 0}{\lim }\frac{\sin \left({\displaystyle \frac{h}{2}}\right)}{\left({\displaystyle \frac{h}{2}}\right)} \\ =\frac{-\cos x}{\sin x\sin x}\times 1 \\ =-cotx\cos ecx \\ Hence,\frac{d}{dx}\left(\cos ecx\right)=-cotx\cos ecx \end{gathered}$$