Find the derivative of the following functions (it is to be under…

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Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): $(\text{x}+\cos\text{x})(\text{x}-\tan\text{x})$

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Let $\text{f(x)}=(\text{x}+\cos\text{x})(\text{x}-\tan\text{x})$ By product rule, $\text{f}'\text{(x)}=(\text{x}+\cos\text{x})\frac{\text{d}}{\text{dx}}(\text{x}-\tan\text{x})+(\text{x}-\tan\text{x})\frac{\text{d}}{\text{dx}}(\text{x}+\cos\text{x})$ $=(\text{x}+\cos\text{x})\Big[\frac{\text{d}}{\text{dx}}(\text{x})-\frac{\text{d}}{\text{dx}}(\tan\text{x})\Big]+(\text{x}-\tan\text{x})(1-\sin\text{x})$ $=(\text{x}+\cos\text{x})\Big[1-\frac{\text{d}}{\text{dx}}\tan\text{x}\Big]+(\text{x}-\tan\text{x})(1-\sin\text{x})$ Let $\text{g(x)}=\tan\text{x.}$ Accordingly, $\text{g(x+h)}=\tan(\text{x+h})$ By first principle,$\text{g}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{g(x+h)}-\text{g(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan\text{(x+h)}-\tan\text{(x)}}{\text{h}}\Big)$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\sin\text{(x+h)}}{\cos\text{x+h}}-\frac{\sin\text{x}}{\cos\text{x}}\Big]$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\sin\text{(x+h)}\cos\text{x}-\sin\text{x}\cos(\text{x+h})}{\cos(\text{x+h})\cos\text{x}}\Big]$ $=\frac{1}{\cos\text{x}}.\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\sin\text{(x+h}-\text{x})}{\cos(\text{x+h})}\Big]$ $=\frac{1}{\cos\text{x}}.\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[\frac{\sin\text{(h})}{\cos(\text{x+h})}\Big]$ $=\frac{1}{\cos\text{x}}.\bigg(\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{(h})}{\text{h}}\bigg).\Bigg(\lim\limits_{\text{h}\rightarrow0}\frac{1}{\cos(\text{x+h})}\Bigg)$ $=\frac{1}{\cos\text{x}}.1.\frac{1}{\cos(\text{x+0})}$ $=\frac{1}{\cos^2\text{x}}$ $=\sec^2\text{x}\ ...(\text{ii})$ Therefore, from (i) and (ii), we obtain $\text{f}'\text{(x)}=(\text{x}+\cos\text{x})(1-\sec^2\text{x})+(\text{x}-\tan\text{x})(1-\sin\text{x})$ $=(\text{x}+\cos\text{x})(-\tan^2\text{x})+(\text{x}-\tan\text{x})(1-\sin\text{x})$ $=-\tan^2\text{x}(\text{x}+\cos\text{x})+(\text{x}-\tan\text{x})(1-\sin\text{x})$

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