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 (√(x+(√(x+(√(x+(√(x+(√(x+...))))))))))=t x+(√(x+(√(x+(√(x+(√(x+...))))))))=t^2 x+t=t^2 t^2 −t−x=0 t=((−(−1)±(√((−1)^2 −4(1)(−x))))/(2(1))) t=((1±(√(1+4x)))/2) y′(x)=(d/dx)(((1±(√(1+4x)))/2)) f ′(a)=lim_(x→a) ((f(x)−f(a))/(x−a)) [First principle /ab−initio Formula] y′(a)=lim_(x→a) (((((1±(√(1+4x)))/2))−(((1±(√(1+4a)))/2)))/(x−a)) =lim_(x→a) ((±(√(1+4x)) ∓ (√(1+4a)))/(2(x−a))) y′(a)=lim_(x→a) (((√(1+4x)) −(√(1+4a)))/(2(x−a))) ∨ y′(a)= lim_(x→a) ((−(√(1+4x)) +(√(1+4a)))/(2(x−a))) =lim_(x→a) ((((√(1+4x)) −(√(1+4a)))/(2(x−a)))×(((√(1+4x)) +(√(1+4a)))/((√(1+4x)) +(√(1+4a))))) =lim_(x→a) ((1+4x−1−4a)/(2(x−a)((√(1+4x)) +(√(1+4a))))) =lim_(x→a) ((4(x−a))/(2(x−a)((√(1+4x)) +(√(1+4a))))) =lim_(x→a) (4/(2((√(1+4x)) +(√(1+4a))))) = (2/((√(1+4a)) + (√(1+4a))))=(1/(√(1+4a))) y′(a)=(1/(√(1+4a))) ⇒y′(x)=(1/(√(1+4x))) y′(a)= lim_(x→a) (((−(√(1+4x)) +(√(1+4a)))/(2(x−a))) ×((+(√(1+4x)) +(√(1+4a)))/(+(√(1+4x)) +(√(1+4a)))) ) =lim_(x→a) (((1+4a−1−4x)/(2(x−a)((√(1+4x)) +(√(1+4a)))))) =lim_(x→a) (((−4(x−a))/(2(x−a)((√(1+4x)) +(√(1+4a)))))) =lim_(x→a) (((−2)/(((√(1+4x)) +(√(1+4a))))))= ((−2)/(((√(1+4a)) +(√(1+4a))))) y′(a) =((−1)/(√(1+4a))) ⇒y′(x)=((−1)/(√(1+4x))) y′(x)=((±1)/(√(1+4x)))](Q1627.png)
$${Let}\:\:\:\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+...}}}}}={t} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}+...}}}}={t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}+{t}={t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{t}^{\mathrm{2}} −{t}−{x}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{t}=\frac{−\left(−\mathrm{1}\right)\pm\sqrt{\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−{x}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:{t}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}}\:\: \\ $$$${y}'\left({x}\right)=\frac{{d}}{{dx}}\left(\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}}\right)\: \\ $$$$\:\:\:\:{f}\:'\left({a}\right)=\underset{{x}\rightarrow{a}} {{lim}}\frac{{f}\left({x}\right)−{f}\left({a}\right)}{{x}−{a}}\:\:\:\:\left[\mathrm{First}\:\mathrm{principle}\:/\mathrm{ab}−\mathrm{initio}\:\mathrm{Formula}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}'\left({a}\right)=\underset{{x}\rightarrow{a}} {{lim}}\frac{\left(\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{x}}}{\mathrm{2}}\right)−\left(\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}}\right)}{{x}−{a}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\frac{\pm\sqrt{\mathrm{1}+\mathrm{4}{x}}\:\:\:\mp\:\:\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}\left({x}−{a}\right)} \\ $$$$\:\:\:\:\:\:{y}'\left({a}\right)=\underset{{x}\rightarrow{a}} {{lim}}\frac{\sqrt{\mathrm{1}+\mathrm{4}{x}}\:−\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}\left({x}−{a}\right)}\:\vee\:{y}'\left({a}\right)=\:\underset{{x}\rightarrow{a}} {{lim}}\frac{−\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}\left({x}−{a}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\left(\frac{\sqrt{\mathrm{1}+\mathrm{4}{x}}\:−\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}\left({x}−{a}\right)}×\frac{\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\:\:\frac{\mathrm{1}+\mathrm{4}{x}−\mathrm{1}−\mathrm{4}{a}}{\mathrm{2}\left({x}−{a}\right)\left(\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\:\:\frac{\mathrm{4}\left({x}−{a}\right)}{\mathrm{2}\left({x}−{a}\right)\left(\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\:\:\frac{\mathrm{4}}{\mathrm{2}\left(\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2}}{\sqrt{\mathrm{1}+\mathrm{4}{a}}\:+\:\sqrt{\mathrm{1}+\mathrm{4}{a}}}=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{4}{a}}} \\ $$$$\:\:\:\:\:\:\:\:\:{y}'\left({a}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{4}{a}}}\:\Rightarrow{y}'\left({x}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{4}{x}}}\:\:\:\:\:\: \\ $$$${y}'\left({a}\right)=\:\underset{{x}\rightarrow{a}} {{lim}}\:\left(\frac{−\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}}{\mathrm{2}\left({x}−{a}\right)}\:×\frac{+\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}}{+\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\left(\frac{\mathrm{1}+\mathrm{4}{a}−\mathrm{1}−\mathrm{4}{x}}{\mathrm{2}\left({x}−{a}\right)\left(\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\left(\frac{−\mathrm{4}\left({x}−{a}\right)}{\mathrm{2}\left({x}−{a}\right)\left(\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\underset{{x}\rightarrow{a}} {{lim}}\left(\frac{−\mathrm{2}}{\left(\sqrt{\mathrm{1}+\mathrm{4}{x}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)}\right)=\:\frac{−\mathrm{2}}{\left(\sqrt{\mathrm{1}+\mathrm{4}{a}}\:+\sqrt{\mathrm{1}+\mathrm{4}{a}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:{y}'\left({a}\right)\:=\frac{−\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{4}{a}}}\:\Rightarrow{y}'\left({x}\right)=\frac{−\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{4}{x}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{y}'\left({x}\right)=\frac{\pm\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{4}{x}}}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
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
$${The}\:{answer}\:{is}\:{in}\:{the}\:{comments} \\ $$$${following}\:{the}\:{question}. \\ $$