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Question Number 4040 by Filup last updated on 27/Dec/15

f (x) = x^{\sin^{x} (x)}

g (x) = x

Why is f^{'} (x) = 0 on g (x) ?

A r e a l l e x t r e m a (\min, \max, inflection)

of f (x) on g (x) ?

Commented by Filup last updated on 27/Dec/15

Commented by Filup last updated on 27/Dec/15

So, when does :

\frac{d f}{d x} = g

Commented by Yozzii last updated on 27/Dec/15

f (x) = x^{{s i n}^{x} x} (*)

I f f (x), x > 0, t a k i n g l n o n b o t h s i d e s o f (*)

\Rightarrow l n f (x) = ({s i n}^{x} x) l n x (* *)

A g a i n, i f l n f (x), ({s i n}^{x} x) l n x > 0,

t a k i n g l n o n b o t h s i d d s o f (* *)

\Rightarrow l n (l n f (x)) = l n {({s i n}^{x} x) l n x}

l n (l n f (x)) = x l n s i n x + l n (l n x) (* * *)

I m p l i c i t l y d i f f e r e n t i a t i n g (* * *) w . r . t x

g i v e s

\frac{\frac{d}{d x} (l n f (x))}{l n f (x)} = 1 \times l n s i n x + x \times \frac{c o s x}{s i n x} + \frac{\frac{d}{d x} (l n x)}{l n x}

\frac{\frac{1}{f (x)} f^{^{'}} (x)}{l n f (x)} = l n s i n x + x c o t x + \frac{1 / x}{l n x}

\frac{f^{^{'}} (x)}{f (x) l n f (x)} = l n s i n x + x c o t x + \frac{1}{x l n x}

f^{^{'}} (x) = f (x) l n f (x) {l n s i n x + x c o t x + \frac{1}{x l n x}}

f^{^{'}} (x) = x^{{s i n}^{x} x} {l n x}^{{s i n}^{x} x} \frac{1}{x l n x} (x l n x {l n s i n x + x c o t x} + 1)

f^{^{'}} (x) = \frac{x^{{s i n}^{x} x} ({s i n}^{x} x) l n x (x (l n x) {l n s i n x + x c o t x} + 1)}{x l n x}

x, l n x > 0

∴ f^{^{'}} (x) = x^{{s i n}^{x} x - 1} ({s i n}^{x} x) (x (l n x) {l n s i n x + x c o t x} + 1)

I f f^{^{'}} (x) = 0

\Rightarrow {s i n}^{x} x = 0 \Rightarrow x = n π, x \in N

f (n π) = {(n π)}^{{(s i n {n π})}^{n π}} = {(n π)}^{0} = 1

I f g (x) = x t o u c h e s f (x) = x^{{s i n}^{x} x}

i t i s n e c e s s a r y t h a t \exists x \in R ∣ f^{^{'}} (x) = g^{^{'}} (x) = 1 .

Commented by Yozzii last updated on 27/Dec/15

Commented by Yozzii last updated on 27/Dec/15

T h e a b o v e g r a p h s a r e g^{^{'}} (x) = 1

a n d y = f^{'} (x) .

W e s e e t h a t i n d e e d \exists x \in R ∣ f^{^{'}} (x) = g^{^{'}} (x) = 1

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

n e i g h b o u r h o o d o f x = n π a t w h i c h

f^{^{'}} (x) = 0 .

Commented by Yozzii last updated on 27/Dec/15

N o t u r n i n g p o i n t s o f y = f (x) l i e

o n y = g (x) a s a c o n s e q u e n c e o f

t h e a b o v e g r a p h s h o w i n g g^{'} (x) = f^{'} (x) = 1 \neq 0

f o r g (x) t o u c h i n g f (x) n e a r x = n π .

Commented by Filup last updated on 27/Dec/15

This is amazing and complicated!

I {don}^{'} t know implicite differentiation,

but this is awesome!!!

Commented by Yozzii last updated on 27/Dec/15

T h i s i s w h y I^{'} m g l a d w e c a n u s e

d i a g r a m s n o w . G r a p h i n g i s i m p o r t a n t .

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