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

Question

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): $\frac{\sin{(\text{x+a)}}}{\cos\text{x}}$

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Answer

Let $\text{f(x)}=\frac{\sin{(\text{x+a})}}{\cos\text{x}}$ By quotient rule, $\text{f}'\text{(x)}=\frac{\cos\text{x}\frac{\text{d}}{\text{dx}}\big[\sin(\text{x+a)}\big]-\sin\text{(x+a)}\frac{\text{d}}{\text{dx}}\cos\text{x}}{\cos^2\text{x}}$ $\text{f}'\text{(x)}=\frac{\cos\text{x}\frac{\text{d}}{\text{dx}}\big[\sin(\text{x+a)}\big]-\sin\text{(x+a)}(-\sin\text{x})}{\cos^2\text{x}}\ ...(\text{i})$ $\text{Let g(x)}=\sin\text{(x+a). Accordingly, (x+h)}=\sin(\text{x+h+a})$ 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}\frac{1}{\text{h}}\big[\sin(\text{x+h+a})-\sin(\text{x+a})\big]$$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[2\cos\Big(\frac{\text{x+h+a+x+a}}{2}\Big)\sin\Big(\frac{\text{x+h+a}-\text{x}-\text{a}}{2}\Big)\Big]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Big[2\cos\Big(\frac{\text{2x+2a+h}}{2}\Big)\sin\Big(\frac{\text{h}}{2}\Big)\Big]$$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\cos\Big(\frac{\text{2x+2a+h}}{2}\Big)\sin\Bigg\{\frac{\Big(\sin\frac{\text{h}}{2}\Big)}{\Big(\frac{\text{h}}{2}\Big)}\Bigg\}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\cos\Big(\frac{\text{2x+2a+h}}{2}\Big)\lim\limits_{\frac{\text{h}}{2}\rightarrow0}\Bigg\{\frac{\Big(\sin\frac{\text{h}}{2}\Big)}{\Big(\frac{\text{h}}{2}\Big)}\Bigg\}$ $\Big[\text{As h}\rightarrow0\Rightarrow\frac{\text{h}}{2}\rightarrow0\Big]$ $=\Big(\cos\frac{\text{2x+2a}}{2}\Big)\times1$ $\Big[\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Big]$$=\cos(\text{x+a})\ ...(\text{ii})$
From (i) and (ii), we obtain $\text{f}'\text{(x)}=\frac{\cos\text{x}.\cos(\text{x+a})+\sin\text{x}\sin(\text{x+a})}{\cos^2\text{x}}$ $=\frac{\cos(\text{x+a}-\text{x})}{\cos^2\text{x}}$ $=\frac{\cos\text{a}}{\cos^2\text{x}}$