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Question Number 5672 by sanusihammed last updated on 23/May/16

D i f f e r e n t i a t e \frac{l n x}{e^{x}} f 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 m e .

Answered by Yozzii last updated on 23/May/16

L e t y = \frac{l n x}{e^{x}} . F r o m f i r s t p r i n c i p l e s

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

= \lim_{h \to 0} \frac{e^{- x - h} l n (x + h) - e^{- x} l n x}{h}

= \lim_{h \to 0} (\frac{l n (x + h)}{{h e}^{x + h}} - \frac{l n x}{{h e}^{x}})

= \lim_{h \to 0} \frac{l n (x + h) - e^{h} l n x}{{h e}^{x} e^{h}}

= e^{- x} \lim_{h \to 0} \frac{1}{{h e}^{h}} (l n (x + h) - {l n x}^{e^{h}})

= e^{- x} \lim_{h \to 0} \frac{1}{{h e}^{h}} l n \frac{x + h}{x^{e^{h}}}

\frac{d y}{d x} = e^{- x} \lim_{h \to 0} l n {(x^{1 - e^{h}} + \frac{h}{x^{e^{h}}})}^{1 / {h e}^{h}}

l n \frac{d y}{d x} = l n {e^{- x}} + l n (\lim_{h \to 0} {\frac{1}{e^{h}} l n [{(x^{1 - e^{h}} + \frac{h}{x^{e^{h}}})}^{1 / h}]})

l n \frac{d y}{d x} = - x + l n (\lim_{h \to 0} \frac{1}{h} {l n x}^{1 - e^{h}} (1 + \frac{h}{x}))

l n \frac{d y}{d x} = - x + l n (\lim_{h \to 0} (\frac{1 - e^{h}}{h} l n x + \frac{1}{h} l n (1 + \frac{h}{x})))

∙ e i s a n u m b e r s u c h t h a t f o r u b e i n g

a n y n u m b e r, e^{u} = 1 + u + \frac{u^{2}}{2!} + \frac{u^{3}}{3!} + \dots + \frac{u^{n}}{n!} + \dots

∙ F o r j \in (- 1, 1] o n e c a n w r i t e

l n (1 + j) = j - \frac{j^{2}}{2} + \frac{j^{3}}{3} - \frac{j^{4}}{4} + \dots + \frac{{(- 1)}^{n + 1} j^{n}}{n} + \dots

∴ l n \frac{d y}{d x} = - x + l n {(l n x) \lim_{h \to 0} \frac{1}{h} (1 - 1 - h - \frac{h^{2}}{2!} - \frac{h^{3}}{3!} - \dots) + \lim_{h \to 0} \frac{1}{h} (\frac{h}{x} - \frac{h^{2}}{2 x^{2}} + \frac{h^{3}}{3 x^{3}} - \dots)}

l n \frac{d y}{d x} = - x + l n {(l n x) \lim_{h \to 0} (- 1 - \frac{h}{2!} - \frac{h^{2}}{3!} - \dots) + \lim_{h \to 0} (\frac{1}{x} - \frac{h}{2 x^{2}} + \frac{h^{2}}{3 x^{3}} - \dots)}

l n \frac{d y}{d x} = - x + l n {(l n x) (- 1) + \frac{1}{x}}

\frac{d y}{d x} = e x p (- x + l n (x^{- 1} - l n x)) = e^{- x} \times (x^{- 1} - l n x)

Commented by sanusihammed last updated on 24/May/16

T h i s i s g r e a t . w o w .

Commented by sanusihammed last updated on 24/May/16

T h a n k s s o m u c h .

Commented by Yozzii last updated on 23/May/16

A l t e r n a t i v e l y, o n e c o u l d p r o v e t h e

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

a n d s u b s e q u e n t l y a p p l y i t .

L e t r (x) = \frac{u (x)}{v (x)} w h e r e u a n d v a r e f u n c t i o n s

o f x a n d v (x) \neq 0 .

∴ r (x + h) = \frac{u (x + h)}{v (x + h)} (h = δ x)

∴ r (x + h) - r (x) = \frac{u (x + h)}{v (x + h)} - \frac{u (x)}{v (x)} = δ r (x)

= \frac{v (x) (u (x + h) - u (x) + u (x)) - u (x) (v (x + h) - v (x) + v (x))}{v (x + h) v (x)}

δ r (x) = \frac{v (x) δ u (x) + v (x) u (x) - u (x) δ v (x) - u (x) v (x)}{v (x + h) v (x)}

δ r (x) = \frac{v (x) δ u (x) - u (x) δ v (x)}{v (x + δ x) v (x)}

B y d e f i n i t i o n o f t h e d e r i v a t i v e,

\frac{d r}{d x} = \lim_{δ x \to 0} \frac{δ r (x)}{δ x} = \lim_{δ x \to 0} \frac{1}{δ x} (\frac{v (x) δ u (x) - u (x) δ v (x)}{v (x + δ x) v (x)})

\frac{d r}{d x} = \lim_{δ x \to 0} \frac{v (x) \frac{δ u (x)}{δ x} - u (x) \frac{δ v (x)}{δ x}}{v (x + δ x) v (x)}

S i n c e v (x) \neq 0 f o r a l l x \in R, b y t h e a l g e b r a

o f l i m i t s,

\Rightarrow \frac{d r}{d x} = \frac{\lim_{δ x \to 0} (v (x) \frac{δ u (x)}{δ x} - u (x) \frac{δ v (x)}{δ x})}{\lim_{δ x \to 0} {v (x + δ x) v (x)}}

\frac{d r}{d x} = \frac{v (x) (\lim_{δ x \to 0} \frac{δ u (x)}{δ x}) - u (x) (\lim_{δ x \to 0} \frac{δ v (x)}{δ x})}{\lim_{δ x \to 0} {v (x + δ x) v (x)}}

N o w, w e s u b s t i t u t e t h e f o l l o w i n g r e s u l t s :

\lim_{δ x \to 0} \frac{δ u (x)}{δ x} = \frac{d u}{d x}, \lim_{δ x \to 0} \frac{δ v (x)}{δ x} = \frac{d v}{d x} &

\lim_{δ x \to 0} {v (x + δ x) v (x)} = v (x + 0) v (x) = v^{2} (x) .

∴ \frac{d r}{d x} = \frac{v (x) \frac{d u}{d x} - u (x) \frac{d v}{d x}}{v^{2} (x)} ◻

S o, i f r (x) = \frac{l n x}{e^{x}}

\Rightarrow \frac{d r}{d x} = \frac{e^{x} x^{- 1} - e^{x} l n x}{e^{2 x}} = \frac{x^{- 1} - l n x}{e^{x}} .

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