Question

Calculate y^{(k)}(0) for 0 <= k <= 5, where y=6 x^4+a x^3 + b x^2 + c x + d (with a,b,c,d the constants)
y^{(0)}(0)=
y^{(1)}(0)=
y^{(2)}(0)=
y^{(3)}(0)=
y^{(4)}(0)=
y^{(5)}(0)=

Answer 1

I know the first one is just d but I can't figure out the rest.

Answer 2

Is y^k the kth derivative?

Answer 3

Is y^1 dy/dx ?
y^2 d^2y / dy^2 ?

Answer 4

or is it just an exponent? y^k = \(y^k\) ?

Answer 5

yeah its just an exponent I think

Answer 6

If the exponent's wrapped in parentheses, they tend to denote orders of derivatives...

Answer 7

I would rather do derivative junk
Looking it as an exponent seems like really mean work

Answer 8

Well they are all evaluated at x = 0

Answer 9

like punishment work

Answer 10

oh that's true lol

Answer 11

then y(0)=d
(y(0))^2=d^2

Answer 12

Is this question from a calc class?

Answer 13

but I would think it means kth derivative

Answer 14

yeah it is

Answer 15

Then it is the derivative.

Answer 16

\[
y=6 x^4+a x^3 + b x^2 + c x + d \\
y^{(0)}(0) = d \\
y^{(1)} = 24x^3 + 3ax^2 + 2bx + c \\
y^{(1)}(0) = c \\
y^{(2)} = ? \\
y^{(2)}(0) = ? \\
....
\]

Answer 17

2b right?

Answer 18

Ok thanks I can get the rest