Results 1 to 4 of 4

Math Help - derivatives help

  1. #1
    Newbie
    Joined
    Nov 2006
    From
    Canada
    Posts
    15

    derivatives help

    I attached a paint file (sorry about the non steady hand) I have no idea how to find the derivative of these.

    derivatives help-derivatives.jpg
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,843
    Thanks
    320
    Awards
    1
    Quote Originally Posted by dcfi6052 View Post
    I attached a paint file (sorry about the non steady hand) I have no idea how to find the derivative of these.
    y = 2 log_7 (5x) + e^{-x^2}

    You can approach this in two ways. First:
    y = log_b(x) => y' = \frac{1}{x ln(b)}

    or simply use the change of base formula:
    log_b(x) = \frac{log_a(x)}{log_a(b)}

    I'm going to use the change of base formula and change the base to e:
    y = 2 log_7 (5x) + e^{-x^2} = 2 \cdot \frac{ln(5x)}{ln(7)} + e^{-x^2}

    Now it's just a matter of using the chain rule:
    y' = \frac{2}{ln(7)} \cdot \frac{1}{5x} \cdot 5 + e^{-x^2} \cdot (-2x)

    y' = \frac{2}{xln(7)} - 2xe^{-x^2}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,843
    Thanks
    320
    Awards
    1
    y = (2x^3 + 5)(3x^2 - x)

    There are two ways to do this.

    First, multiply out the expression:
    y = 6x^5 - 2x^4 + 15x^2 - 5x

    Then
    y' = 30x^4 - 8x^3 + 30x - 5

    Or use the product rule:
    y = (2x^3 + 5)(3x^2 - x)

    y' = (6x^2)(3x^2 - x) + (2x^3 + 5)(6x - 1)

    which reduces to the same expression as above.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,843
    Thanks
    320
    Awards
    1
    y = \frac{x^2 - 7}{2x+1} \cdot e^{x^2}

    This is a combination of the quotient and product rules:
    y' = \frac{(2x)(2x + 1) - (x^2 - 7)(2)}{(2x + 1)^2} \cdot e^{x^2} + \frac{x^2 - 7}{2x + 1} \cdot e^{x^2} \cdot (2x)

    y' = \frac{2x^2 + 2x + 14}{(2x + 1)^2} \cdot e^{x^2} + \frac{2x(x^2 - 7)}{2x + 1} \cdot e^{x^2}

    I'll let you do the rest of the simplifying.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivatives and Anti-Derivatives
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 6th 2011, 06:21 AM
  2. Replies: 1
    Last Post: July 19th 2010, 04:09 PM
  3. Derivatives with both a and y
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 4th 2009, 09:17 AM
  4. Replies: 4
    Last Post: February 10th 2009, 09:54 PM
  5. Trig derivatives/anti-derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 10th 2009, 01:34 PM

Search Tags


/mathhelpforum @mathhelpforum