Results 1 to 12 of 12

Math Help - Very cool

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    79

    Very cool

    D (x) = (x + 1) (x + 2) ^ 2 (x +3) ^ 3 (x + 4) ^ 4 (x + 5)

    Find the derivative at -1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by dapore View Post
    D (x) = (x + 1) (x + 2) ^ 2 (x +3) ^ 3 (x + 4) ^ 4 (x + 5)

    Find the derivative at -1
    Use log differentiation

    ln[D(x)] = ln[(x+1)(x+2)^2(x+3)^3(x+4)^4(x+5)]  = ln(x+1) + 2ln(x+2) + 3ln(x+3) + 4ln(x+4) + ln(x+5)


    Recall that due the chain rule: \frac{d}{dx}ln[f(x)] = \frac{f'(x)}{f(x)}


    Actually, looking at it I don't think it is differentiable at x=-1 because of the 1/(x+1) term
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,700
    Thanks
    454
    Quote Originally Posted by dapore View Post
    D (x) = (x + 1) (x + 2) ^ 2 (x +3) ^ 3 (x + 4) ^ 4 (x + 5)

    Find the derivative at -1
    D(x) = f(x) \cdot g(x) \cdot h(x) \cdot p(x) \cdot q(x)<br />

    D'(x) = f'(x) \cdot g(x) \cdot h(x) \cdot p(x) \cdot q(x) + f(x) \cdot g'(x) \cdot h(x) \cdot p(x) \cdot q(x)  + f(x) \cdot g(x) \cdot h'(x) \cdot p(x) \cdot q(x) + f(x) \cdot g(x) \cdot h(x) \cdot p'(x) \cdot q(x) + f(x) \cdot g(x) \cdot h(x) \cdot p(x) \cdot q'(x)

    since f(x) = (x+1) , and f(-1) = 0 ...

    D'(-1) = (1)(1^2)(2^3)(3^4)(4)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42
    You could go right to the definition

     <br />
D'(-1) = \lim_{x \to -1} \frac{D(x) - D(-1)}{x+1} = \lim_{x \to -1} \frac{(x+1)(x+2)^2(x+3)^3(x+4)^4(x+5)}{x+1} <br />

    = \lim_{x \to -1} (x+2)^2(x+3)^3(x+4)^4(x+5) = 1^2 \cdot 2^3 \cdot 3^4 \cdot 4 as Skeeter said.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2008
    Posts
    249
    I'm curious, why didn't logarithmic differentiation work for this problem?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by oblixps View Post
    I'm curious, why didn't logarithmic differentiation work for this problem?
    I was just about to say that.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,700
    Thanks
    454
    Quote Originally Posted by oblixps View Post
    I'm curious, why didn't logarithmic differentiation work for this problem?
    actually, it does ... take the limit as x \to e^{i \pi} of the log expression posted by e^(i*pi)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Aug 2008
    Posts
    249
    but ln(1+e^(i*pi)) = ln(0), doesn't the limit not exist?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Aug 2008
    Posts
    249
    Quote Originally Posted by e^(i*pi) View Post
    Use log differentiation

    ln[D(x)] = ln[(x+1)(x+2)^2(x+3)^3(x+4)^4(x+5)]  = ln(x+1) + 2ln(x+2) + 3ln(x+3) + 4ln(x+4) + ln(x+5)


    Recall that due the chain rule: \frac{d}{dx}ln[f(x)] = \frac{f'(x)}{f(x)}


    Actually, looking at it I don't think it is differentiable at x=-1 because of the 1/(x+1) term
    after differentiating both sides and multiplying both sides by (x+1)(x+2)^2(x+3)^3(x+4)^4(x+5), you'll get y'= [\frac{1}{x+1} + \frac{2}{x+2}+\frac{3}{x+3}+\frac{4}{x+4}+\frac{1}  {x+5}]  [(x+1)(x+2)^2(x+3)^3(x+4)^4(x+5)]

    then distribute and plug in x = -1 and you'll see that all the terms will cancel out and become zero except for the first term, since the (x+1)s cancel out. when you plug in x = -1 to what you have left, you'll get (1^2)(2^3)(3^4)(4)
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Dec 2009
    Posts
    79
    You the most beautiful tribute
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Jul 2009
    Posts
    192
    Thanks
    4
    Quote Originally Posted by oblixps View Post
    after differentiating both sides and multiplying both sides by (x+1)(x+2)^2(x+3)^3(x+4)^4(x+5), you'll get y'= [\frac{1}{x+1} + \frac{2}{x+2}+\frac{3}{x+3}+\frac{4}{x+4}+\frac{1}  {x+5}]  [(x+1)(x+2)^2(x+3)^3(x+4)^4(x+5)]

    then distribute and plug in x = -1 and you'll see that all the terms will cancel out and become zero except for the first term, since the (x+1)s cancel out. when you plug in x = -1 to what you have left, you'll get (1^2)(2^3)(3^4)(4)
    ye so whats this about there being a problem with 1/(1+x)? am i missing something
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Aug 2008
    Posts
    249
    e^(i*pi) said the derivative does not exist because of the 1/(x+1) term but in fact it just cancels out when you distribute so the derivative does exist. I was just correcting a small mistake.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cool forum! I need a little help....
    Posted in the Statistics Forum
    Replies: 2
    Last Post: February 20th 2010, 01:57 PM
  2. A few cool ones here
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: March 26th 2009, 05:00 PM
  3. cool integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: December 29th 2008, 12:07 PM
  4. cool series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 4th 2008, 05:53 AM
  5. cool theorem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 14th 2008, 03:50 PM

Search Tags


/mathhelpforum @mathhelpforum