Question Number 1624 by 112358 last updated on 27/Aug/15
$${Find}\:{the}\:{first}\:{derivative}\:{of} \\ $$$${y}\left({x}\right)=\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+…}}}}} \\ $$$${from}\:{first}\:{principles}.\: \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 27/Aug/15
$${Let}\:\:\:\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+…}}}}}={t} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+…}}}}={t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}+{t}={t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}^{\mathrm{2}} −{t}−{x}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{t}=\frac{−\left(−\mathrm{1}\right)\pm\sqrt{\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−{x}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:{t}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}}\:\: \\ $$$${y}'\left({x}\right)=\frac{{d}}{{dx}}\left(\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}}\right)\: \\ $$$$\:\:\:\:{f}\:'\left({a}\right)=\underset{{x}\rightarrow{a}} {{lim}}\frac{{f}\left({x}\right)−{f}\left({a}\right)}{{x}−{a}}\:\:\:\:\left[\mathrm{First}\:\mathrm{principle}\:/\mathrm{ab}−\mathrm{initio}\:\mathrm{Formula}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}'\left({a}\right)=\underset{{x}\rightarrow{a}} {{lim}}\frac{\left(\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}}\right)−\left(\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}}\right)}{{x}−{a}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\frac{\pm\sqrt{\mathrm{1}+\mathrm{4}{x}}\:\:\:\mp\:\:\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}\left({x}−{a}\right)} \\ $$$$\:\:\:\:\:\:{y}'\left({a}\right)=\underset{{x}\rightarrow{a}} {{lim}}\frac{\sqrt{\mathrm{1}+\mathrm{4}{x}}\:−\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}\left({x}−{a}\right)}\:\vee\:{y}'\left({a}\right)=\:\underset{{x}\rightarrow{a}} {{lim}}\frac{−\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}\left({x}−{a}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\left(\frac{\sqrt{\mathrm{1}+\mathrm{4}{x}}\:−\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}\left({x}−{a}\right)}×\frac{\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\:\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\:\:\frac{\mathrm{1}+\mathrm{4}{x}−\mathrm{1}−\mathrm{4}{a}}{\mathrm{2}\left({x}−{a}\right)\left(\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\:\:\frac{\mathrm{4}\left({x}−{a}\right)}{\mathrm{2}\left({x}−{a}\right)\left(\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\:\:\frac{\mathrm{4}}{\mathrm{2}\left(\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}+\mathrm{4}{a}}\:+\:\sqrt{\mathrm{1}+\mathrm{4}{a}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{a}}} \\ $$$$\:\:\:\:\:\:\:\:\:{y}'\left({a}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{a}}}\:\Rightarrow{y}'\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{x}}}\:\:\:\:\:\: \\ $$$${y}'\left({a}\right)=\:\underset{{x}\rightarrow{a}} {{lim}}\:\left(\frac{−\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}\left({x}−{a}\right)}\:×\frac{+\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}}{+\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\left(\frac{\mathrm{1}+\mathrm{4}{a}−\mathrm{1}−\mathrm{4}{x}}{\mathrm{2}\left({x}−{a}\right)\left(\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\left(\frac{−\mathrm{4}\left({x}−{a}\right)}{\mathrm{2}\left({x}−{a}\right)\left(\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\left(\frac{−\mathrm{2}}{\left(\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)}\right)=\:\frac{−\mathrm{2}}{\left(\sqrt{\mathrm{1}+\mathrm{4}{a}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:{y}'\left({a}\right)\:=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{a}}}\:\Rightarrow{y}'\left({x}\right)=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{x}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{y}'\left({x}\right)=\frac{\pm\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{4}{x}}}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Commented by 112358 last updated on 27/Aug/15
$${If}\:−{x}\:{is}\:{constant}\:{with}\:{t}\:{variable} \\ $$$${shouldn}'{t}\:{it}\:{be}\: \\ $$$$\:\:\:\:\:\:\:{t}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}}\:? \\ $$
Commented by Rasheed Soomro last updated on 28/Aug/15
$$\mathrm{Y}{es},\:{it}\:{is}\:{a}\:{calculation}\:{mistake}!\:{And}\:{I}\:{am}\:{going}\:{to}\:{correct}\:{it}. \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{mentioning}. \\ $$
Commented by Yozzian last updated on 27/Aug/15
$${Looks}\:{correct}\:{sir}!\:{My}\:{approach} \\ $$$${took}\:{the}\:{form} \\ $$$$\:\:\:\:\:\frac{{dy}}{{dx}}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{f}\left({x}+{h}\right)−{f}\left({x}\right)}{{h}} \\ $$
Commented by Rasheed Soomro last updated on 28/Aug/15
$${Thanks}! \\ $$
Answered by Yozzian last updated on 27/Aug/15
$${The}\:{answer}\:{is}\:{in}\:{the}\:{comments} \\ $$$${following}\:{the}\:{question}. \\ $$