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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
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

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Zarkon Group TitleBest ResponseYou've already chosen the best response.1
only one of the answers makes sense
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
a) has an x in it...that's not good
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
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)
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
b) and c) will give answers that are not integers
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
Thats true lol
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
te derivative of y evaluated at zero will clearly be an integer
 2 years ago

ujjwal Group TitleBest ResponseYou've already chosen the best response.0
How do you know that the derivative of y evaluated at zero will be an integer?
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
Have you done any of the cases I mentioned to convince yourself?
 2 years ago

ujjwal Group TitleBest ResponseYou've already chosen the best response.0
I haven't done such problems before..
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
only one answer will work
 2 years ago

ujjwal Group TitleBest ResponseYou've already chosen the best response.0
@Zarkon , you mean to say that we can put value n=1 and see which option satisfies.. And that's it..??
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.1
Remember the product of integers will be integers and the sum of integers will be integers
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
if they give you options...which they do...yes
 2 years ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
at least for this problem
 2 years ago

ujjwal Group TitleBest ResponseYou've already chosen the best response.0
Thanks a lot @Zarkon and @freckles ...
 2 years ago
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