Results 1 to 2 of 2

Math Help - Derivative calculations

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    9

    Derivative calculations

    Hi could someone please help me differentiate?

    <br />
r(t) = ln(2t\sqrt(t+1))<br />

    I was thinking of using the product rule.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by studentme View Post
    Hi could someone please help me differentiate?

    <br />
r(t) = ln(2t\sqrt(t+1))<br />

    I was thinking of using the product rule.
    It would come after the chain rule.

    [u(v(x))]'=v'(x)u'(v(x))

    Here, u(y)=\ln(y) and v(t)=2t \sqrt{t+1} and since you'll have to compute v'(t), this is where you will have to use the product rule.

    Otherwise, you can write it this way :
    r(t)=\ln(2t \sqrt{t+1})=\ln(2t)+\ln(\sqrt{t+1})=\ln(2)+\ln(t)+  \ln((t+1) {1/2}) =\ln(2)+\ln(t)+\frac 12 \ln(t+1)
    ln(2) is a constant, hence its derivative is 0.
    ln(t) has a well-known derivative.
    ln(t+1) has a simple derivative you can compute by using the chain rule.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Vector Calculations
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: August 21st 2011, 03:37 AM
  2. basic calculations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 17th 2010, 05:31 AM
  3. calculations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 4th 2010, 12:41 PM
  4. LOG and EXP Calculations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 14th 2009, 11:39 AM
  5. Calculations?
    Posted in the Business Math Forum
    Replies: 3
    Last Post: January 29th 2008, 06:40 AM

Search Tags


/mathhelpforum @mathhelpforum