Differentiate the following from first principle: 2x^2-1 - Vidyadip

Question

Differentiate the following from first principle:$\sqrt{2\text{x}^2-1}$

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Answer

We have, $\text{f(x)}=\sqrt{2\text{x}^2+1}$ $\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x+h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{2(\text{x+h})^2+1}-\sqrt{2\text{x}^2+1}}{\text{h}}$ Multiplying Numerator and denominator by ${\sqrt{2(\text{x+h})^2+1}+\sqrt{2\text{x}^2+1}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\bigg\{2(\text{x+h})^2+1-(2\text{x}^2+1)\bigg\}}{\text{h}\bigg(\sqrt{2(\text{x+h})^2+1}+\sqrt{2\text{x}^2+1}\bigg)}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{2\text{x}^2+2\text{h}^2+4\text{xh}+1-2\text{x}^2-1}{\text{h}\bigg(\sqrt{2(\text{x+h})^2+1}+\sqrt{2\text{x}^2+1}\bigg)}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{4\text{xh}+2\text{h}^2}{\text{h}\sqrt{2(\text{x+h})^2+1}+\sqrt{2\text{x}^2+1}}$ $=\frac{4\text{x}}{2\sqrt{2\text{x}^2+1}}$ $=\frac{2\text{x}}{\sqrt{2\text{x}^2+1}}$ $$

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