Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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+\(\frac{1}{2}\)+\(\frac{1}{3}\)+.... +\(\frac{1}{n}\))
 one year ago
 one year 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+\(\frac{1}{2}\)+\(\frac{1}{3}\)+.... +\(\frac{1}{n}\))
 one year ago
 one year ago

This Question is Closed

ZarkonBest ResponseYou've already chosen the best response.1
only one of the answers makes sense
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
a) has an x in it...that's not good
 one year ago

frecklesBest 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)
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
b) and c) will give answers that are not integers
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
te derivative of y evaluated at zero will clearly be an integer
 one year ago

ujjwalBest ResponseYou've already chosen the best response.0
How do you know that the derivative of y evaluated at zero will be an integer?
 one year ago

frecklesBest ResponseYou've already chosen the best response.1
Have you done any of the cases I mentioned to convince yourself?
 one year ago

ujjwalBest ResponseYou've already chosen the best response.0
I haven't done such problems before..
 one year ago

frecklesBest 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
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
only one answer will work
 one year ago

ujjwalBest 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..??
 one year ago

frecklesBest ResponseYou've already chosen the best response.1
Remember the product of integers will be integers and the sum of integers will be integers
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
if they give you options...which they do...yes
 one year ago

ZarkonBest ResponseYou've already chosen the best response.1
at least for this problem
 one year ago

ujjwalBest ResponseYou've already chosen the best response.0
Thanks a lot @Zarkon and @freckles ...
 one year ago
See more questions >>>
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.