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Derivatives — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDerivatives5 Marks
Question
Differentiate the following from first principle$\sqrt{\tan\text{x}}$
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Answer

We have,$\text{f}(\text{x})=\sqrt{\tan\text{x}}$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\tan(\text{x}+\text{h})}-\sqrt{\tan\text{x}}}{\text{h}}$
Multiplying Numerator and Denominator by $\sqrt{\tan(\text{x}+\text{h})}+\sqrt{\tan\text{x}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\tan(\text{x}+\text{h})-\tan{\text{x}}}{\text{h}\big(\sqrt{\tan(\text{x}+\text{h})}+\sqrt{\tan\text{x}}\big)}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{x}+\text{h}-\text{x})}{\text{h}.\cos(\text{x}+\text{h})\cos\text{x}\big[\sqrt{\tan(\text{x}+\text{h})}+\sqrt{\tan\text{x}}\big]}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}\times\frac{1}{\cos(\text{x}+\text{h})\cos\text{x}\big[\sqrt{\tan(\text{x}+\text{h})}+\sqrt{\tan\text{x}}\big]}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{1}{\cos^2\text{x}.2\sqrt{\tan\text{x}}}\ \Bigg[\lim_\limits{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Bigg]$
$=\frac{1}{2}\frac{\sec^2\text{x}}{\sqrt{\tan\text{x}}}\ \Bigg[\frac{1}{\cos^2\text{x}}=\sec^2\text{x}\Bigg]$

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