Differentiate the following from first principle: 1 3-x - Vidyadip

Question

Differentiate the following from first principle:

$\frac{1}{\sqrt{3-\text{x}}}$

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Answer

We have,

$\text{f(x)}=\frac{1}{\sqrt{{3}-\text{x}}}$

$\because\text{f}'\text{(x)}=\lim\limits_{\text{h}\rightarrow0}\frac{\text{f(x+h)}-\text{f(x)}}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\frac{1}{\sqrt{3-(\text{x+h)}}}-\frac{1}{\sqrt{3-\text{x}}}}{\text{h}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{3-\text{x}}-\sqrt{3-\text{(x+h)}}}{\sqrt{3-\text{x}}{\sqrt{3-\text{(x+h)}}\times\text{h}}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{3-\text{x}}-\sqrt{3-\text{(x+h)}}}{\sqrt{3-\text{x}}\sqrt{3-\text{(x+h)}\text{h}}}\times\frac{\sqrt{3-\text{x}}+\sqrt{3-\text{(x+h)}}}{\sqrt{3-\text{x}}+\sqrt{3-\text{(x+h)}}}$

$\Big[ $Rationalising the numerator by $\sqrt{3-\text{x}}+\sqrt{3-\text{(x+h)}}\Big]$

$=\lim\limits_{\text{h}\rightarrow0}\frac{(3-\text{x})-\big(3-\text{(x+h)}\big)}{\sqrt{3-\text{x}}{\sqrt{3-\text{(x+h)}}\times\text{h}\bigg(\sqrt{3-\text{x}}+\sqrt{3-\text{(x+h)}}\bigg)}}$

$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}}{\sqrt{3-\text{x}}{\sqrt{3-\text{(x+h)}}\times\text{h}\big(\sqrt{3-\text{x}}+{\sqrt{3-\text{(x+h)}}}\big)}}$

$=\frac{1}{(3-\text{x})\times2\sqrt{3-\text{x}}}$

$=\frac{1}{2(3-\text{x})^\frac{3}{2}}$

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