What a jerk!

Calculus Level 5

x x x . . . \LARGE x^{x^{x^{.^{.^.}}}}

A jerk is defined to be the third derivative of a position. Let f ( x ) f(x) denote the function of the infinite power tower as described above. Evaluate the jerk of this function at x = 1 x = 1 .


Inspiration .


The answer is 9.

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3 solutions

Ayush Verma
Jul 30, 2015

y = x y l n y = y l n x d i f f e r e n t i a t i n g w . r . t x y 1 y = y 1 l n x + y x y 1 = y 1 ( y l n x ) + y 2 x = y 1 l n y + y 2 x . . . ( i ) d i f f e r e n t i a t e a g a i n , y 2 = y 2 l n y + ( y 1 ) 2 y + 2 x y y 1 y 2 x 2 . . . ( i i ) d i f f e r e n t i a t e a g a i n , y 3 = ( y 3 l n y + y 2 y 1 y ) + 2 y y 1 y 2 ( y 1 ) 3 y 2 + x 2 { 2 y y 1 + 2 x ( y 1 ) 2 + 2 x y y 2 2 y y 1 } 2 x ( 2 x y y 1 y 2 ) ( x 2 ) 2 . . . ( i i i ) i f x = 1 y ( 1 ) = 1 B y e q n ( i ) , y 1 ( 1 ) = y 1 ( 1 ) l n 1 + 1 2 1 = 1 B y e q n ( i i ) , y 2 ( 1 ) = y 2 ( 1 ) l n 1 + ( 1 ) 2 1 + 2 × 1 × 1 × 1 1 2 1 2 = 2 B y e q n ( i i ) , y 3 ( 1 ) = ( y 3 ( 1 ) l n 1 + 2 × 1 1 ) + 2 × 1 × 1 × 2 ( 1 ) 3 1 2 + 1 2 { 2 × 1 × 1 + 2 × 1 × 1 2 + 2 × 1 × 1 × 2 2 × 1 × 1 } 2 × 1 ( 2 × 1 × 1 × 1 1 2 ) ( 1 2 ) 2 = 2 + 4 1 1 + ( 2 + 2 + 4 2 ) 2 ( 2 1 ) 1 = 9 y={ x }^{ y }\Rightarrow lny=ylnx\\ differentiating\quad w.r.t\quad x\\ \Rightarrow \cfrac { { y }_{ 1 } }{ y } ={ y }_{ 1 }lnx+\cfrac { y }{ x } \\ \\ \Rightarrow { y }_{ 1 }={ y }_{ 1 }\left( { y }lnx \right) +\cfrac { { y }^{ 2 } }{ x } ={ y }_{ 1 }lny+\cfrac { { y }^{ 2 } }{ x } \quad \quad ...\left( i \right) \\ \\ differentiate\quad again,\\ \\ \Rightarrow { y }_{ 2 }={ y }_{ 2 }lny+\cfrac { { \left( { y }_{ 1 } \right) }^{ 2 } }{ y } +\cfrac { 2xy{ y }_{ 1 }-{ y }^{ 2 } }{ { x }^{ 2 } } \quad \quad ...\left( ii \right) \\ \\ differentiate\quad again,\\ \\ \Rightarrow { y }_{ 3 }=\left( { y }_{ 3 }lny+{ y }_{ 2 }\cfrac { { y }_{ 1 } }{ y } \right) +\cfrac { 2y{ y }_{ 1 }{ y }_{ 2 }-{ \left( { y }_{ 1 } \right) }^{ 3 } }{ { y }^{ 2 } } \\ \\ +\cfrac { { x }^{ 2 }\left\{ 2y{ y }_{ 1 }+2x{ \left( { y }_{ 1 } \right) }^{ 2 }+2xy{ y }_{ 2 }-2y{ y }_{ 1 } \right\} -2x\left( 2xy{ y }_{ 1 }-{ y }^{ 2 } \right) }{ { \left( { x }^{ 2 } \right) }^{ 2 } } ...\left( iii \right) \\ \\ if\quad x=1\Rightarrow y\left( 1 \right) =1\\ \\ By\quad { eq }^{ n }\quad \left( i \right) ,\\ \\ { y }_{ 1 }\left( 1 \right) ={ y }_{ 1 }\left( 1 \right) ln1+\cfrac { { 1 }^{ 2 } }{ 1 } =1\\ \\ By\quad { eq }^{ n }\quad \left( ii \right) ,\\ \\ { y }_{ 2 }\left( 1 \right) ={ y }_{ 2 }\left( 1 \right) ln1+\cfrac { { \left( 1 \right) }^{ 2 } }{ 1 } +\cfrac { 2\times 1\times 1\times 1-{ 1 }^{ 2 } }{ { 1 }^{ 2 } } =2\\ \\ By\quad { eq }^{ n }\quad \left( ii \right) ,\\ \\ { y }_{ 3 }\left( 1 \right) =\left( { y }_{ 3 }\left( 1 \right) ln1+2\times \cfrac { 1 }{ 1 } \right) +\cfrac { 2\times 1\times 1\times 2-{ \left( 1 \right) }^{ 3 } }{ { 1 }^{ 2 } } \\ \\ +\cfrac { { 1 }^{ 2 }\left\{ 2\times 1\times 1+2\times 1\times 1^{ 2 }+2\times 1\times 1\times 2-2\times 1\times 1 \right\} -2\times 1\left( 2\times 1\times 1\times 1-{ 1 }^{ 2 } \right) }{ { \left( { 1 }^{ 2 } \right) }^{ 2 } } \\ \\ =2+\cfrac { 4-1 }{ 1 } +\cfrac { \left( 2+2+4-2 \right) -2\left( 2-1 \right) }{ 1 } =9

Hadia Qadir
Aug 11, 2015

he third derivative of the function listed at x = 1 is equal to 9.

what are you saying hadia ???

Muhammad Aftab Alam - 5 years, 10 months ago

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First of all. It's important to define the limit "x^x^x^x^..." precisely, for there are ambiguities. For instance (3^3)^3=27^3=3^9 =\= 3^27=3^(3^3) Adopting the first interpretation: x^x^x^x^...=lim exp(x^n.logx)=1, if x<=1 (and diverges, if x>1).

Hadia Qadir - 5 years, 10 months ago

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thanks .. hadia ....still,,

Muhammad Aftab Alam - 5 years, 10 months ago
Andriane Casuga
Aug 10, 2015

First of all. It's important to define the limit "x^x^x^x^..." precisely, for there are ambiguities. For instance (3^3)^3=27^3=3^9 =\= 3^27=3^(3^3) Adopting the first interpretation: x^x^x^x^...=lim exp(x^n.logx)=1, if x<=1 (and diverges, if x>1). This derivative should be 0 at 1, therefore.

Nope. There is no ambiguities. You should always perform the operations from the top to bottom. See this .

Pi Han Goh - 5 years, 10 months ago

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Sorry, Im just a teenager

andriane casuga - 5 years, 10 months ago

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No need to apologize. =D

Pi Han Goh - 5 years, 10 months ago

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@Pi Han Goh Ok, thanks

andriane casuga - 5 years, 10 months ago

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