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If f(x)=|x^nn !2cosxcos(npi)/2 4sinxsin(...

If `f(x)=|x^nn !2cosxcos(npi)/2 4sinxsin(npi)/2 8|` then find the value of `(d^n)/(dx^n)([f(x)])_(x=0)dot(n in z)dot`

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The correct Answer is:
0

`(d^(n))/(dx^(n))[f(x)]=|{:((d^(n))/(dx^(n))(x^(n)),n!,2),((d^(n))/(dx^(n))(cos x),cos (npi)/(2),4),((d^(n))/(dx^(n))(sin x), sin (npi)/(2),8):}|`
`=|{:(n!,n!,2),(cos(x+(npi)/(2)),cos""(npi)/(2),4),(sin(x+(npi)/(2)),sin""(npi)/(2),8):}|`
`"or "(d^(n))/(dx^(n))[f(x)]_(x=0)=0`
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