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ujjwal

  • 3 years ago

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+\(\frac{1}{2}\)+\(\frac{1}{3}\)+.... +\(\frac{1}{n}\))

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  1. Zarkon
    • 3 years ago
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    only one of the answers makes sense

  2. Zarkon
    • 3 years ago
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    a) has an x in it...that's not good

  3. freckles
    • 3 years ago
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    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
    • 3 years ago
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    b) and c) will give answers that are not integers

  5. freckles
    • 3 years ago
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    Thats true lol

  6. Zarkon
    • 3 years ago
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    te derivative of y evaluated at zero will clearly be an integer

  7. ujjwal
    • 3 years ago
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    How do you know that the derivative of y evaluated at zero will be an integer?

  8. freckles
    • 3 years ago
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    Have you done any of the cases I mentioned to convince yourself?

  9. ujjwal
    • 3 years ago
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    I haven't done such problems before..

  10. freckles
    • 3 years ago
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    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
    • 3 years ago
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    let n=1

  12. Zarkon
    • 3 years ago
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    only one answer will work

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

  14. freckles
    • 3 years ago
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    Remember the product of integers will be integers and the sum of integers will be integers

  15. Zarkon
    • 3 years ago
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    if they give you options...which they do...yes

  16. Zarkon
    • 3 years ago
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    at least for this problem

  17. ujjwal
    • 3 years ago
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    Thanks a lot @Zarkon and @freckles ...

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