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Calculus Level 5

f ( x ) = e 2 sin 1 ( x ) \large f(x) = e^{2\sin^{-1}(x)}

Find the 10th derivative of f ( x ) f(x) above at x = 0 x = 0 .


The answer is 1740800.

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1 solution

Vighnesh Raut
Aug 17, 2015

L e t y = e 2 sin 1 x Let\quad y={ e }^{ 2\sin ^{ -1 }{ x } }

Now , differentiating with respect to x x , we get

y = 2 y 1 x 2 ( 1 x 2 ) y = 2 y \\ { y }'=\frac { 2y }{ \sqrt { 1-{ x }^{ 2 } } } \\ \left( \sqrt { 1-{ x }^{ 2 } } \right) { y }'=2y

Again differentiating with respect to x x and simplifying the euation, we get

( 1 x 2 ) y x y = 4 y \left( 1-{ x }^{ 2 } \right) y''-xy'=4y ------------equation 1 1

Here, we use the Leibnitz's rule of finding n t h nth derivative of product of two functions:

( u v ) n = ( n 0 ) u v n + ( n 1 ) u 1 v n 1 + ( n 2 ) u 2 v n 2 + . . . . . . + ( n n ) u n v { \left( uv \right) }_{ n }=\left( \begin{matrix} n \\ 0 \end{matrix} \right) { u }{ v }_{ n }+\left( \begin{matrix} n \\ 1 \end{matrix} \right) { u }_{ 1 }{ v }_{ n-1 }+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { u }_{ 2 }{ v }_{ n-2 }+......+\left( \begin{matrix} n \\ n \end{matrix} \right) { u }_{ n }{ v }

where x n { x }_{ n } is the n t h nth derivative of x x and ( a b ) = a ! b ! ( a b ) ! \left( \begin{matrix} a \\ b \end{matrix} \right) =\frac { a! }{ b!(a-b)! }

Now, applying Leibnitz's rule to equation 1 1 , we get

( 1 x 2 ) y n + 2 2 n x y n + 1 n ( n 1 ) y n x y n + 1 n y n = 4 y n \left( 1-{ x }^{ 2 } \right) { y }_{ n+2 }-2nx{ y }_{ n+1 }-n\left( n-1 \right) { y }_{ n }-x{ y }_{ n+1 }-n{ y }_{ n }=4{ y }_{ n }

On simplifying, we get

( 1 x 2 ) y n + 2 ( 2 n + 1 ) x y n + 1 ( 4 + n 2 ) y n = 0 \left( 1-{ x }^{ 2 } \right) { y }_{ n+2 }-(2n+1)x{ y }_{ n+1 }-(4+{ n }^{ 2 }){ y }_{ n }=0 -------------equation 2 2

where y n { y }_{ n } is the n t h nth derivative of y y . Now, substituting value of x x as 0 0 in equation 2 2 , we get

y n + 2 = ( 4 + n 2 ) y n { y }_{ n+2 }=\left( 4+{ n }^{ 2 } \right) { y }_{ n } ---------equation 3 3

We know that, y ( 0 ) = 1 { y }\left( 0 \right) =1

y 1 ( 0 ) = 2 { y }_{ 1 }\left( 0 \right) =2

Using equation 3 3 , relation for y n ( 0 ) { y }_{ n }\left( 0 \right) becomes ,

y n ( 0 ) = 2 ( 2 2 + 1 2 ) ( 2 2 + 3 2 ) . . . . . ( 2 2 + ( n 2 ) 2 ) , n i s o d d y n ( 0 ) = 2 2 ( 2 2 + 2 2 ) ( 2 2 + 4 2 ) . . . . . ( 2 2 + ( n 2 ) 2 ) , n i s e v e n { y }_{ n }\left( 0 \right) =2\left( { 2 }^{ 2 }+{ 1 }^{ 2 } \right) \left( { 2 }^{ 2 }+{ 3 }^{ 2 } \right) .....\left( { 2 }^{ 2 }+{ \left( n-2 \right) }^{ 2 } \right) ,\quad n\quad is\quad odd\\ { y }_{ n }\left( 0 \right) ={ 2 }^{ 2 }\left( { 2 }^{ 2 }+{ 2 }^{ 2 } \right) \left( { 2 }^{ 2 }+{ 4 }^{ 2 } \right) .....\left( { 2 }^{ 2 }+{ \left( n-2 \right) }^{ 2 } \right) ,\quad n\quad is\quad even

Substituting value of n = 10 n=10 , we get y 10 ( 0 ) = 1740800 { y }_{ 10 }\left( 0 \right) =1740800

A slightly easier approach is to find the Maclaurin series expansion directly, though it may not yield as nice a formula that you found for y n y_n .

Calvin Lin Staff - 3 years, 7 months ago

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Indeed! Finding the Maclaurin series would definitely save the time. The above procedure is a complete waste of time if the result is the only goal. My intention here was to generate a result , that would work for any value of 'n'.

Vighnesh Raut - 3 years, 6 months ago

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