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Derivatives — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsDerivatives4 Marks
Question
Differentiate the following from the first principle$\sqrt{\sin(\text{3x}+1)}$
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Answer

We have, $\text{f}(\text{x})=\sqrt{\sin(\text{3x}+1)}$
$\because\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\sin3(\text{x}+\text{h})+1}-\sqrt{\sin(\text{3x}+1)}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\sin(\text{3x}+\text{3h})+1}-\sqrt{\sin(\text{3x}+1)}}{\text{h}}\times\frac{{\sqrt{\sin(\text{3x}+\text{3h})+1}+\sqrt{\sin(\text{3x}+1)}}}{{\sqrt{\sin(\text{3x}+\text{3h})+1}+\sqrt{\sin(\text{3x}+1)}}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{3x}+\text{3h}+1)-\sin(\text{3x}+1)}{\text{h}\big(\sqrt{\sin(\text{3x}+\text{3h})+1}+\sqrt{\sin(\text{3x}+1)}\big)}$
$=\lim_\limits{\text{h}\rightarrow0}2\cos\Big(\text{3x}+1+\frac{\text{3h}}{2}\Big)\times\frac{\sin\frac{\text{3h}}{2}}{\frac{\text{3h}}{2}}\times\frac{3}{2}\times\frac{1}{\sqrt{\sin(\text{3x}+\text{3h})+1}+\sqrt{\sin(\text{3x}+1)}}$
$=\frac{3\cos(\text{3x}+1)}{2\sqrt{\sin(\text{3x}+1)}}$ $\Bigg[\lim_\limits{\text{h}\rightarrow0}\frac{\sin\frac{\text{3h}}{2}}{\frac{\text{3h}}{2}}=1\Bigg]$

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