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Derivatives — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSDerivatives4 Marks
Question
Differentiate the following from first principle$\tan(\text{2x}+1)$
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Answer

$\text{f}(\text{x})=\tan(\text{2x}+1)$$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan\big\{2(\text{x}+\text{h})+1\big\}-\tan(\text{2x}+1)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{2h}+1-\text{2x}-1)}{\text{h}.\cos\big\{2(\text{x}+\text{h})+1\big\}\cos(\text{2x}+1)}$
$\Big[\because\tan\text{A}-\tan\text{B}=\frac{\sin(\text{A}-\text{B})}{\cos\text{A}.\cos\text{B}}\Big]$
$=\lim_\limits{\text{h}\rightarrow0}\frac{2\sin2\text{h}}{2\text{h}.\cos(\text{2x}+\text{2h}+1)\cos(\text{2x}+1)}$
Multiplying both, numerator and denominator by 2.
$\therefore\lim_\limits{\text{h}\rightarrow0}2\Big(\frac{\sin2\text{h}}{2}\Big)\times\frac{1}{\cos(\text{2x}+\text{2h}+1)\cos(\text{2x}+1)}$
$=\frac{2}{\cos^2(\text{2x}+1)}\ \Big[\because\lim_\limits{\text{h}\rightarrow0}\frac{\sin2\text{h}}{2}=1\Big]$
$=2\sec^2(\text{2x}+1)\ \Big[\because\sec^2\text{x}=\frac{1}{\cos^2\text{x}}\Big]$
$=2\sec^ 2(\text{2x}+1)$

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