Find the derivative of the following functions:

Question

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

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Answer

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