ujjwal Group Title 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+$$\frac{1}{2}$$+$$\frac{1}{3}$$+.... +$$\frac{1}{n}$$) 2 years ago 2 years ago

1. Zarkon Group Title

only one of the answers makes sense

2. Zarkon Group Title

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

3. freckles Group Title

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 Group Title

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

5. freckles Group Title

Thats true lol

6. Zarkon Group Title

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

7. ujjwal Group Title

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

8. freckles Group Title

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

9. ujjwal Group Title

I haven't done such problems before..

10. freckles Group Title

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 Group Title

let n=1

12. Zarkon Group Title

only one answer will work

13. ujjwal Group Title

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

14. freckles Group Title

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

15. Zarkon Group Title

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

16. Zarkon Group Title

at least for this problem

17. ujjwal Group Title

Thanks a lot @Zarkon and @freckles ...