Question

If y=(x+1)(x+2)(x+3)......(x+n) then dy/dx at x=0 is:
a)n! ln(1+x)
b)n! ln2
c)n! ln3
d)n! (1+\(\frac{1}{2}\)+\(\frac{1}{3}\)+.... +\(\frac{1}{n}\))

Answer 1

only one of the answers makes sense

Answer 2

a) has an x in it...that's not good

Answer 3

You can do it for like small values of n and see if you see one of the patterns above
like for n=1,n=2,n=3,n=4
I would stop at n=4
(also hint: don't multiply the junk out after applying product rule)

Answer 4

b) and c) will give answers that are not integers

Answer 5

Thats true lol

Answer 6

te derivative of y evaluated at zero will clearly be an integer

Answer 7

How do you know that the derivative of y evaluated at zero will be an integer?

Answer 8

Have you done any of the cases I mentioned to convince yourself?

Answer 9

I haven't done such problems before..

Answer 10

If y=x+1, then y'=1 so y'(at x=0) =1
If y=(x+1)(x+2), then y'=? <---use product rule....after using product rule don't multiply the junk out and then evaluate y' at x=0 and see what you get

Answer 11

let n=1

Answer 12

only one answer will work

Answer 13

@Zarkon , you mean to say that we can put value n=1 and see which option satisfies.. And that's it..??

Answer 14

Remember the product of integers will be integers and the sum of integers will be integers

Answer 15

if they give you options...which they do...yes

Answer 16

at least for this problem

Answer 17

Thanks a lot @Zarkon and @freckles ...