Differentiate the following from first principle 2x - Vidyadip

Question

Differentiate the following from first principle$\sin\sqrt{2\text{x}}$

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Answer

We have,$\text{f}(\text{x})=\sin\sqrt{2\text{x}}$
$\text{f}'(\text{x})=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sqrt{\sin2(\text{x}+\text{h})}-\sin\sqrt{2\text{x}}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{2\sin\Big(\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{2}\Big)\cos\Big(\frac{\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}}}{2}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Bigg(\frac{\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}}}{2}\Bigg)(\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}})(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}})}{\Bigg(\frac{(\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}})}{2}\Bigg)(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}})\text{h}}\cos\Bigg(\frac{(\sqrt{2(\text{x}+\text{h})}+\sqrt{2\text{x}})}{2}\Bigg)$$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin\Bigg(\frac{(\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}})}{2}\Bigg)}{\Bigg(\frac{(\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}})}{2}\Bigg)}\lim_\limits{\text{h}\rightarrow0}\frac{{2(\text{x}+\text{h})}-{2\text{x}}}{(\sqrt{2(\text{x}+\text{h})}-\sqrt{2\text{x}})\text{h}}\lim_\limits{\text{h}\rightarrow0}\cos\Bigg(\frac{\sqrt{2(\text{x}+\text{h})+\text{2x}}}{2}\Bigg)$
$=1\times\frac{2}{2\sqrt{2\text{x}}}\cos(\sqrt{2\text{x}})$
$=\frac{\cos(\sqrt{2\text{x}})}{\sqrt{2\text{x}}}$

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