Find the derivative of the following function from first principl… - Vidyadip

Question

Find the derivative of the following function from first principle. $\frac{\text{x}+1}{\text{x}-1}$

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Answer

Let $\text{f}(\text{x})=\frac{\text{x}+1}{\text{x}-1}$. Accordingly, from the first principle, $\text{f}'(\text{x})=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f}(\text{x}+\text{h})-\text{f}(\text{x})}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{\bigg(\frac{\text{x}+\text{h}+1}{\text{x}+\text{h}-1}-\frac{\text{x}+1}{\text{x}-1}\bigg)}{\text{h}}$ $=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{(\text{x}-1)(\text{x}+\text{h}+1)-(\text{x}+1)(\text{x}+\text{h}-1)}{(\text{x}-1)(\text{x}+\text{h}-1)}\Bigg]$$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{(\text{x}^2+\text{hx}+\text{x}-\text{x}-\text{h}-1)(\text{x}^2+\text{hx}-\text{x}+\text{x}+\text{h}-1}{(\text{x}+1)(\text{x}+\text{h}-1)}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{-2\text{h}}{(\text{x}-1)(\text{x}+\text{h})-1}\Bigg]$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{h}}\Bigg[\frac{-2}{(\text{x}-1)(\text{x}+\text{h})-1}\Bigg]$ $=\frac{2}{(\text{x}-1)(\text{x}-1)}=\frac{-2}{(\text{x}-1)^2}$

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