Find the derivative of the following functions: cosecx

Question

Find the derivative of the following functions: $\text{cosecx}$

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Answer

Let $\text{f}(\text{x}) = 5\sec\text{x}+4\cos\text{x}$. Accordingly, from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}[\text{cosec}(\text{x}+\text{h})-\text{cosec}\text{x}]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{1}{\sin(\text{x}+\text{h})}-\frac{1}{\sin\text{x}}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{\sin\text{x}-\sin(\text{x}+\text{h})}{\sin(\text{x}+\text{h})\sin\text{x}}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{2\cos\bigg(\frac{\text{x}+\text{x}+\text{h}}{2}\bigg)\sin\bigg (\frac{\text{x}-\text{x}-\text{h}}{2}\bigg)}{\sin(\text{x}+\text{h})\sin\text{x}}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}\frac{1}{\text{h}}\bigg[\frac{2\cos\bigg(\frac{2\text{x}+\text{h}}{2}\bigg)\sin\bigg (-\frac{\text{h}}{2}\bigg)}{\sin(\text{x}+\text{h})\sin\text{x}}\bigg]$ $=\lim\limits_{{\text{h}}\rightarrow0}\Bigg[\frac{-\cos\bigg(\frac{2\text{x}+\text{h}}{2}\bigg).\frac{\sin\bigg (\frac{\text{h}}{2}\bigg)}{\frac{\text{h}}{2}}}{\sin(\text{x}+\text{h})\sin\text{x}}\Bigg]$ $=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{-\cos\big(\frac{2\text{x}+\text{h}}{2}\big)}{\sin(\text{x}+\text{h})\sin\text {x}}\Bigg).\lim\limits_{\text{h}\rightarrow0}\frac{\sin\big(\frac{\text{h}}{2}\big)}{\big(\frac{\text{h}}{2}\big)}$ $=\bigg(\frac{-\cos\text{x}}{\sin\text{x}\sin\text{x}}\bigg).1$