Question Number 35933 by mondodotto@gmail.com last updated on 26/May/18
$$\boldsymbol{\mathrm{differentiate}}\:\boldsymbol{\mathrm{from}} \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{first}}\:\boldsymbol{\mathrm{principle}} \\ $$$$\boldsymbol{\mathrm{y}}=\frac{\mathrm{1}}{\sqrt{\boldsymbol{{x}}}} \\ $$
Answered by ajfour last updated on 26/May/18
![dy=(1/(√(x+dx)))−(1/(√x)) =(((√x)−(√(x+dx)))/((√x)(√(x+dx)))) = (((√(x[))1−(1+(dx/(2x)))])/(∣x∣)) (dy/dx) = −(1/(2x(√x))) .](Q35935.png)
$${dy}=\frac{\mathrm{1}}{\sqrt{{x}+{dx}}}−\frac{\mathrm{1}}{\sqrt{{x}}} \\ $$$$\:\:\:\:=\frac{\sqrt{{x}}−\sqrt{{x}+{dx}}}{\sqrt{{x}}\sqrt{{x}+{dx}}}\:=\:\frac{\left.\sqrt{{x}\left[\right.}\mathrm{1}−\left(\mathrm{1}+\frac{{dx}}{\mathrm{2}{x}}\right)\right]}{\mid{x}\mid} \\ $$$$\:\frac{{dy}}{{dx}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}{x}\sqrt{{x}}}\:. \\ $$