Find-the-first-derivative-of-y-x-x-x-x-x-x-from-first-principles-

Question Number 1624 by 112358 last updated on 27/Aug/15

F i n d t h e f i r s t d e r i v a t i v e o f

y (x) = \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \dots}}}}}

f r o m f i r s t p r i n c i p l e s .

Commented by Rasheed Soomro last updated on 27/Aug/15

L e t \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \dots}}}}} = t

x + \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{x + \dots}}}} = t^{2}

x + t = t^{2}

t^{2} - t - x = 0

t = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 (1) (- x)}}{2 (1)}

t = \frac{1 \pm \sqrt{1 + 4 x}}{2}

y^{'} (x) = \frac{d}{d x} (\frac{1 \pm \sqrt{1 + 4 x}}{2})

f^{'} (a) = \underset{x \to a}{l i m} \frac{f (x) - f (a)}{x - a} [First principle / ab - initio Formula]

y^{'} (a) = \underset{x \to a}{l i m} \frac{(\frac{1 \pm \sqrt{1 + 4 x}}{2}) - (\frac{1 \pm \sqrt{1 + 4 a}}{2})}{x - a}

= \underset{x \to a}{l i m} \frac{\pm \sqrt{1 + 4 x} \mp \sqrt{1 + 4 a}}{2 (x - a)}

y^{'} (a) = \underset{x \to a}{l i m} \frac{\sqrt{1 + 4 x} - \sqrt{1 + 4 a}}{2 (x - a)} \lor y^{'} (a) = \underset{x \to a}{l i m} \frac{- \sqrt{1 + 4 x} + \sqrt{1 + 4 a}}{2 (x - a)}

= \underset{x \to a}{l i m} (\frac{\sqrt{1 + 4 x} - \sqrt{1 + 4 a}}{2 (x - a)} \times \frac{\sqrt{1 + 4 x} + \sqrt{1 + 4 a}}{\sqrt{1 + 4 x} + \sqrt{1 + 4 a}})

= \underset{x \to a}{l i m} \frac{1 + 4 x - 1 - 4 a}{2 (x - a) (\sqrt{1 + 4 x} + \sqrt{1 + 4 a})}

= \underset{x \to a}{l i m} \frac{4 (x - a)}{2 (x - a) (\sqrt{1 + 4 x} + \sqrt{1 + 4 a})}

= \underset{x \to a}{l i m} \frac{4}{2 (\sqrt{1 + 4 x} + \sqrt{1 + 4 a})}

= \frac{2}{\sqrt{1 + 4 a} + \sqrt{1 + 4 a}} = \frac{1}{\sqrt{1 + 4 a}}

y^{'} (a) = \frac{1}{\sqrt{1 + 4 a}} \Rightarrow y^{'} (x) = \frac{1}{\sqrt{1 + 4 x}}

y^{'} (a) = \underset{x \to a}{l i m} (\frac{- \sqrt{1 + 4 x} + \sqrt{1 + 4 a}}{2 (x - a)} \times \frac{+ \sqrt{1 + 4 x} + \sqrt{1 + 4 a}}{+ \sqrt{1 + 4 x} + \sqrt{1 + 4 a}})

= \underset{x \to a}{l i m} (\frac{1 + 4 a - 1 - 4 x}{2 (x - a) (\sqrt{1 + 4 x} + \sqrt{1 + 4 a})})

= \underset{x \to a}{l i m} (\frac{- 4 (x - a)}{2 (x - a) (\sqrt{1 + 4 x} + \sqrt{1 + 4 a})})

= \underset{x \to a}{l i m} (\frac{- 2}{(\sqrt{1 + 4 x} + \sqrt{1 + 4 a})}) = \frac{- 2}{(\sqrt{1 + 4 a} + \sqrt{1 + 4 a})}

y^{'} (a) = \frac{- 1}{\sqrt{1 + 4 a}} \Rightarrow y^{'} (x) = \frac{- 1}{\sqrt{1 + 4 x}}

y^{'} (x) = \frac{\pm 1}{\sqrt{1 + 4 x}}

Commented by 112358 last updated on 27/Aug/15

I f - x i s c o n s t a n t w i t h t v a r i a b l e

{s h o u l d n}^{'} t i t b e

t = \frac{1 \pm \sqrt{1 + 4 x}}{2} ?

Commented by Rasheed Soomro last updated on 28/Aug/15

Y e s, i t i s a c a l c u l a t i o n m i s t a k e! A n d I a m g o i n g t o c o r r e c t i t .

Thanks for mentioning .

Commented by Yozzian last updated on 27/Aug/15

L o o k s c o r r e c t s i r! M y a p p r o a c h

t o o k t h e f o r m

\frac{d y}{d x} = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

Commented by Rasheed Soomro last updated on 28/Aug/15

T h a n k s!

Answered by Yozzian last updated on 27/Aug/15

T h e a n s w e r i s i n t h e c o m m e n t s

f o l l o w i n g t h e q u e s t i o n .

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