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Question Number 35933 by mondodotto@gmail.com last updated on 26/May/18

$$\mathbf{differentiate}\mathbf{from}$$$$\mathbf{the}\mathbf{first}\mathbf{principle}$$$$\mathbf{y}=\frac{1}{\sqrt{\mathit{x}}}$$

Answered by ajfour last updated on 26/May/18

$$dy=\frac{1}{\sqrt{x+dx}}-\frac{1}{\sqrt{x}}$$$$=\frac{\sqrt{x}-\sqrt{x+dx}}{\sqrt{x}\sqrt{x+dx}}=\frac{\sqrt{x[}1-(1+\frac{dx}{2x})]}{\mid x\mid }$$$$\frac{dy}{dx}=-\frac{1}{2x\sqrt{x}}.$$

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