## ujjwal 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+$$\frac{1}{2}$$+$$\frac{1}{3}$$+.... +$$\frac{1}{n}$$) one year ago one year ago

1. Zarkon

only one of the answers makes sense

2. Zarkon

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

3. freckles

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)

4. Zarkon

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

5. freckles

Thats true lol

6. Zarkon

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

7. ujjwal

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

8. freckles

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

9. ujjwal

I haven't done such problems before..

10. freckles

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

11. Zarkon

let n=1

12. Zarkon

13. ujjwal

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

14. freckles

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

15. Zarkon

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

16. Zarkon

at least for this problem

17. ujjwal

Thanks a lot @Zarkon and @freckles ...