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

Mathematics
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only one of the answers makes sense
a) has an x in it...that's not good
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)

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b) and c) will give answers that are not integers
Thats true lol
te derivative of y evaluated at zero will clearly be an integer
How do you know that the derivative of y evaluated at zero will be an integer?
Have you done any of the cases I mentioned to convince yourself?
I haven't done such problems before..
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
let n=1
only one answer will work
@Zarkon , you mean to say that we can put value n=1 and see which option satisfies.. And that's it..??
Remember the product of integers will be integers and the sum of integers will be integers
if they give you options...which they do...yes
at least for this problem
Thanks a lot @Zarkon and @freckles ...

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