A community for students.
Here's the question you clicked on:
 0 viewing
ujjwal
 4 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+\(\frac{1}{2}\)+\(\frac{1}{3}\)+.... +\(\frac{1}{n}\))
ujjwal
 4 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+\(\frac{1}{2}\)+\(\frac{1}{3}\)+.... +\(\frac{1}{n}\))

This Question is Closed

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.1only one of the answers makes sense

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

freckles
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1b) and c) will give answers that are not integers

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

ujjwal
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1Have you done any of the cases I mentioned to convince yourself?

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

freckles
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1only one answer will work

ujjwal
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.1if they give you options...which they do...yes

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

ujjwal
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks a lot @Zarkon and @freckles ...
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.