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Question Number 5715 by sanusihammed last updated on 24/May/16

D i f f e r e n t i a t e x^{x} f r o m t h e f i r s t p r i n c i p l e .

p l e a s e h e l p

Answered by Yozzii last updated on 25/May/16

L e t y = x^{x} . I f x > 0 w e c a n w r i t e

l n y = {l n x}^{x} = x l n x . K e e p i n m i n d t h a t

y i s a f u n c t i o n o f x . T o d i f f e r e n t i a t e

t h e e q u a t i o n w i t h r e s p e c t t o x, w e n e e d t o f i n d

\frac{d}{d x} (l n y) a n d \frac{d}{d x} (x l n x) .

L e t l n y = u . T h e n, b y f i r s t p r i n c i p l e s,

\frac{d u}{d y} = \lim_{δ y \to 0} \frac{δ u}{δ y} = \lim_{δ y \to 0} \frac{l n (y + δ y) - l n y}{δ y}

\frac{d u}{d y} = \lim_{δ y \to 0} \frac{l n (1 + \frac{δ y}{y})}{δ y} = \lim_{δ y \to 0} l n {(1 + \frac{δ y}{y})}^{1 / δ y}

L e t k = δ y / y \Rightarrow δ y = y k

\frac{d u}{d y} = \lim_{k \to 0} {(1 + u)}^{1 / (y k)} = \frac{1}{y} {\lim_{k \to 0} l n {(1 + k)}^{1 / k}}

l e t k = 1 / n .

∴ \frac{d u}{d y} = \frac{1}{y} {\lim_{n \to \infty} l n {(1 + \frac{1}{n})}^{n}} = \frac{1}{y} l n {\lim_{n \to \infty} {(1 + n^{- 1})}^{n}} = \frac{1}{y} l n e

\Rightarrow \frac{d u}{d y} = 1 / y \dots \dots . (1)

N o w, \frac{d u}{d x} = \frac{d u}{d y} \times \frac{d y}{d x} a c c o r d i n g t o t h e c h a i n r u l e .

P R O O F : F r o m f i r s t p r i n c i p l e s

\frac{d u}{d x} = \lim_{δ x \to 0} \frac{δ u}{δ x} = \lim_{δ x \to 0} (\frac{δ u}{δ y} \times \frac{δ y}{δ x}) = (\lim_{δ x \to 0} \frac{δ u}{δ y}) (\lim_{δ x \to 0} \frac{δ y}{δ x})

I f y = f (x) s u c h t h a t δ y = f (x + δ x) - f (x)

\Rightarrow a s δ x \to 0, f (x + δ x) - f (x) \to 0 o r δ y \to 0 .

∴ \lim_{δ x \to 0} \frac{δ u}{δ y} = \lim_{δ y \to 0} \frac{δ u}{δ y} = \frac{d u}{d y} a n d \lim_{δ x \to 0} \frac{δ y}{δ x} = \frac{d y}{d x} .

\Rightarrow \frac{d u}{d x} = \frac{d u}{d y} \times \frac{d y}{d x} .

∴ \frac{d u}{d x} = \frac{d (l n y)}{d x} = \frac{d (l n y)}{d y} \times \frac{d y}{d x} = \frac{1}{y} ∙ \frac{d y}{d x} .

N o w,

\frac{d (x l n x)}{d x} = \lim_{δ x \to 0} \frac{(x + δ x) l n (x + δ x) - x l n x}{δ x}

= \lim_{δ x \to 0} \frac{x l n (x + δ x) - x l n x + δ x l n (x + δ x)}{δ x}

= \lim_{δ x \to 0} {\frac{x}{δ x} l n (1 + \frac{δ x}{x}) + l n (x + δ x)}

= x {\lim_{δ x \to 0} \frac{1}{δ x} l n (1 + \frac{δ x}{x})} + l n (x + 0)

W e j u s t f o u n d t h e r e s u l t o f \lim_{δ y \to 0} \frac{1}{δ y} l n (1 + \frac{δ y}{y}) = \frac{1}{y} f r o m (1) .

H e n c e, s i m i l a r l y \lim_{δ x \to 0} \frac{1}{δ x} l n (1 + \frac{δ x}{x}) = \frac{1}{x} .

\Rightarrow \frac{d (x l n x)}{d x} = x \times \frac{1}{x} + l n x = 1 + l n x (x > 0)

∴ \frac{1}{y} \times \frac{d y}{d x} = 1 + l n x \Rightarrow \frac{d y}{d x} = y (1 + l n x) = x^{x} (1 + l n x)

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