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y-x-Find-dy-dx-by-first-principle-




Question Number 162675 by Mathematification last updated on 31/Dec/21
y = (√x)  Find   (dy/dx)  by first principle.
$${y}\:=\:\sqrt{{x}} \\ $$$${Find}\:\:\:\frac{{dy}}{{dx}}\:\:{by}\:{first}\:{principle}. \\ $$
Answered by tounghoungko last updated on 31/Dec/21
 (dy/dx) = lim_(h→0)  ((f(x+h)−f(x))/h)         = lim_(h→0)  (((√(x+h))−(√x))/h) = lim_(h→0)  (h/(h ((√(x+h))+(√x))))        = lim_(h→0)  (1/( (√(x+h))+(√x))) = (1/(2(√x)))
$$\:\frac{{dy}}{{dx}}\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}} \\ $$$$\:\:\:\:\:\:\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}+{h}}−\sqrt{{x}}}{{h}}\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{h}}{{h}\:\left(\sqrt{{x}+{h}}+\sqrt{{x}}\right)} \\ $$$$\:\:\:\:\:\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{{x}+{h}}+\sqrt{{x}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\: \\ $$

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