Question

Given g(x) = x^99 + 100x^15 + 1557x^12 + 11548x^5 + 123456, what is the 100th derivative of g(x)?

Answer 1

`0`

Explanation:

`"Each time we derive a positive integer power of x with respect"`
`"to x, the exponent decreases by 1, until we have only a"`
`"constant left and if we derive a constant we get 0."`

`f(x) = a x^n " (n positive integer) "=> f'(x) = a*n*x^(n-1)`

`"If we derive f(x) k times we get"`

`f^(k) (x) = a n (n-1) ...(n-k+1) x^(n-k)`

`"if k = n we get "x^0 = 1" and "`

`f^(n)(x) = a * n! = " constant (not dependent on x anymore)"`

`"If we derive a constant we get 0 and if we derive 0, it remains"`
`"0 so"`
`"if k > n we get 0."`

`"And here k > n for all terms so we get"`

`0 + 0 + ... + 0 = 0`