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

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