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Derivatives — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDerivatives5 Marks
Question
Differentiate the following from the first principle$\sqrt{\sin\text{2x}}$
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Answer

We have,$\text{f}(\text{x})=\sqrt{\sin\text{2x}}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\sin2(\text{x}+\text{h})}-\sqrt{\sin\text{2x}}}{\text{h}}$
Multiplying numerator and denominator by $\Big(\sqrt{\sin2(\text{x}+\text{h})}+\sqrt{\sin\text{2x}}\Big)$
$\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\sin2(\text{x}+\text{h})}-\sqrt{\sin\text{2x}}}{\text{h}}\times\frac{\sqrt{\sin2(\text{x}+\text{h})}+\sqrt{\sin\text{2x}}}{\sqrt{\sin2(\text{x}+\text{h})}+\sqrt{\sin2\text{x}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{2x}+\text{2h})-\sin\text{2x}}{\Big(\sqrt{\sin(\text{2x}+\text{2h})}+\sqrt{\sin2\text{x}}\Big)}$ $\Big[\sin\text{c}-\sin\text{d}=2\cos\frac{\text{c}+\text{d}}{2}\sin\frac{\text{c}-\text{d}}{2}\Big]$
$\lim_\limits{\text{h}\rightarrow0}\frac{2\cos(\text{2x}+\text{h})\times\sin\text{h}}{\text{h}}\times\frac{1}{\sqrt{\sin(\text{2x}+\text{2h})}+\sqrt{\sin\text{2x}}}$
$=\frac{2\cos\text{2x}}{2\sqrt{\sin\text{2x}}}$
$=\frac{\cos\text{2x}}{\sqrt{\sin\text{2x}}}$

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