Question
Differentiate the following from first principle$\tan\text{2x}$
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Answer
$\text{f}(\text{x})=\tan\text{2x}$$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan2(\text{x}+\text{h})-\tan2\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{2h}-\text{2x})}{\text{h}.\cos(\text{2x}+\text{2h})\cos\text{2x}}\ \Big[\because\tan\text{A}-\tan\text{B}=\frac{\sin\text{A}-\text{B}}{\cos\text{A}.\cos{\text{B}}}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{\text{h}.\cos(\text{2x}+\text{2h})\cos2\text{x}}$
$=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin2\text{h}}{2\text{h}}\Big)\times\frac{1\times2}{\cos(\text{2h}+\text{2x})\cos2\text{x}}$
$=\frac{2}{\cos2\text{x}.\cos2\text{x}}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{2\text{h}}=1\Big]$
$=2\sec^22\text{x}\ \Big[\because\frac{1}{\cos^2\text{x}}=\sec^2\text{x}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan2(\text{x}+\text{h})-\tan2\text{x}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{2h}-\text{2x})}{\text{h}.\cos(\text{2x}+\text{2h})\cos\text{2x}}\ \Big[\because\tan\text{A}-\tan\text{B}=\frac{\sin\text{A}-\text{B}}{\cos\text{A}.\cos{\text{B}}}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{\text{h}.\cos(\text{2x}+\text{2h})\cos2\text{x}}$
$=\lim_\limits{\text{h}\rightarrow0}\Big(\frac{\sin2\text{h}}{2\text{h}}\Big)\times\frac{1\times2}{\cos(\text{2h}+\text{2x})\cos2\text{x}}$
$=\frac{2}{\cos2\text{x}.\cos2\text{x}}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{2\text{h}}=1\Big]$
$=2\sec^22\text{x}\ \Big[\because\frac{1}{\cos^2\text{x}}=\sec^2\text{x}\Big]$
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