Question
Differentiate the following from first principle:$\frac{1}{\sqrt{\text{x}}}$
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Answer
We have,$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{\sqrt{\text{x+h}}}-\frac{1}{\sqrt{\text{x}}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{\text{x}}-\sqrt{\text{x+h}}}{\sqrt{\text{x}}\sqrt{\text{x+h}}.\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{\text{x}}-\sqrt{\text{x+h}}}{\sqrt{\text{x}}\sqrt{\text{x+h}}.\text{h}}\times\frac{\sqrt{\text{x}}+\sqrt{\text{x+h}}}{\sqrt{\text{x}}+\sqrt{\text{x+h}}.\text{h}}$
$$$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x}-\text{(x+h)}}{\sqrt{\text{x}}\sqrt{\text{x+h}}.\text{h}\big(\sqrt{\text{x}}+\sqrt{\text{x+h}}\big)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{\sqrt{\text{x}}\sqrt{\text{x+h}}.\text{h}\big(\sqrt{\text{x}}+\sqrt{\text{x+h}}\big)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{\text{x}\sqrt{\text{x+h}}+\sqrt{\text{x}}\big(\sqrt{\text{x+h}}\big)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{2\text{x}\sqrt{\text{x}}}$
$=\frac{-1}{2}\text{x}^{-\frac{3}{2}}$$$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{\sqrt{\text{x+h}}}-\frac{1}{\sqrt{\text{x}}}}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{\text{x}}-\sqrt{\text{x+h}}}{\sqrt{\text{x}}\sqrt{\text{x+h}}.\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{\text{x}}-\sqrt{\text{x+h}}}{\sqrt{\text{x}}\sqrt{\text{x+h}}.\text{h}}\times\frac{\sqrt{\text{x}}+\sqrt{\text{x+h}}}{\sqrt{\text{x}}+\sqrt{\text{x+h}}.\text{h}}$
$$$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{x}-\text{(x+h)}}{\sqrt{\text{x}}\sqrt{\text{x+h}}.\text{h}\big(\sqrt{\text{x}}+\sqrt{\text{x+h}}\big)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-\text{h}}{\sqrt{\text{x}}\sqrt{\text{x+h}}.\text{h}\big(\sqrt{\text{x}}+\sqrt{\text{x+h}}\big)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{\text{x}\sqrt{\text{x+h}}+\sqrt{\text{x}}\big(\sqrt{\text{x+h}}\big)}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-1}{2\text{x}\sqrt{\text{x}}}$
$=\frac{-1}{2}\text{x}^{-\frac{3}{2}}$$$
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