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 2 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+\(\frac{1}{2}\)+\(\frac{1}{3}\)+.... +\(\frac{1}{n}\))
 2 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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Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1only one of the answers makes sense

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

freckles
 2 years ago
Best ResponseYou've already chosen the best response.1You 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)

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

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

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

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

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

freckles
 2 years ago
Best ResponseYou've already chosen the best response.1If 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

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

ujjwal
 2 years ago
Best 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..??

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

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

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

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