Question

f

Answer 1

@hartnn

Answer 2

@dpaInc

Answer 3

E

Answer 4

Can you explain why?

Answer 5

oh, but there is that f'(1)...
generally the derivative would use the power rule to get bnx^(n-1) and the index of summation would move up by one.... let me think about the rest.

Answer 6

Yes, I still think E.

Answer 7

\[\frac{d}{dx}(b _{n}x ^{n})=b _{n}nx ^{(n-1)}\]When you expand the initial summation given, the first term is a constant. When you take the derviative term for term, the derivative of the zero term is zero. That shifts the index of summation up by one. Substitute x=1 into the result and you get (E).

Answer 8

Oh yeah

Answer 9

That made a lot of sense. THanks

Answer 10

Cool! That was something I learned in diff eq. they really ask you nasty questions for a calculus test.

Answer 11

Oh this isn't even the test.. the tests are rough

Answer 12

also, i am picking all the questions i dont know how to do ive done like 30 other ones not that bad

Answer 13

Re-post in case someone is faster than me.

Answer 14

ok