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Question Number 118887 by rexfordattacudjoe last updated on 20/Oct/20

Answered by benjo_mathlover last updated on 20/Oct/20

 f ′(x) = lim_(h→0)  (((√(x+h)) −(√x))/h)   = lim_(h→0)  ((x+h−x)/h) × lim_(h→0)  (1/( (√(x+h)) + (√x) ))  = 1×(1/(2(√x))) = (1/(2(√x)))

$$\:{f}\:'\left({x}\right)\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{{x}+{h}}\:−\sqrt{{x}}}{{h}} \\ $$$$\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}+{h}−{x}}{{h}}\:×\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{{x}+{h}}\:+\:\sqrt{{x}}\:} \\ $$$$=\:\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}} \\ $$

Answered by ebi last updated on 20/Oct/20

f^′ (x)=lim_(h→0)  ((f(x+h)−f(x))/h)  f^′ (x)=lim_(h→0)  (((√(x+h))−(√x))/h)  f^′ (x)=lim_(h→0)  (((√(x+h))−(√x))/h)×(((√(x+h))+(√x))/( (√(x+h))+(√x)))  f^′ (x)=lim_(h→0)  (h/( h((√(x+h))+(√x))))  f^′ (x)=(1/( (√(x+0))+(√x)))  f^′ (x)=(1/( 2(√x)))

$${f}^{'} \left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{li}{m}}\:\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}} \\ $$$${f}^{'} \left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{li}{m}}\:\frac{\sqrt{{x}+{h}}−\sqrt{{x}}}{{h}} \\ $$$${f}^{'} \left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{li}{m}}\:\frac{\sqrt{{x}+{h}}−\sqrt{{x}}}{{h}}×\frac{\sqrt{{x}+{h}}+\sqrt{{x}}}{\:\sqrt{{x}+{h}}+\sqrt{{x}}} \\ $$$${f}^{'} \left({x}\right)=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{li}{m}}\:\frac{{h}}{\:{h}\left(\sqrt{{x}+{h}}+\sqrt{{x}}\right)} \\ $$$${f}^{'} \left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{{x}+\mathrm{0}}+\sqrt{{x}}} \\ $$$${f}^{'} \left({x}\right)=\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{{x}}} \\ $$

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