Results 1 to 7 of 7

Math Help - Derivatives

  1. #1
    Newbie
    Joined
    Aug 2007
    Posts
    22

    Question Derivatives

    Hello would gratefully appreciate help on the following question, also would you be able to guid me through how you got the answers
    thank you
    Amanda x
    Find dy/dx of the following:

    1) y=3^(3x-1) + (1-2x)In(1-2x)

    2) y= 2x-1 -log4(x+1)
    3x+4

    3)
    y= e^x (2x+1)^1/2

    4) intergrate and differentiate:
    xIn(x+1)
    Last edited by Amanda-UK; August 15th 2007 at 04:48 AM. Reason: question 2's layout was wrong as first part is a fraction
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Amanda-UK View Post
    Find dy/dx of the following:

    1) y=3^(3x-1) + (1-2x)In(1-2x)
    I presume that "In(1 - 2x)" is "ln(1 - 2x)"?

    Note that \frac{d}{dx}ln(x) = \frac{1}{x}.
    Also \frac{d}{dx}a^x = ln(a) \cdot a^x.

    Thus, using the chain rule:
    y = 3^{3x - 1} + (1 - 2x)ln(1 - 2x)

    \frac{dy}{dx} = \left ( ln(3x - 1) \cdot 3^{3x - 1} \right ) \cdot 3 + (-2) \cdot ln(1 - 2x) + (1 - 2x) \cdot \left ( \frac{1}{1 - 2x} \right ) \cdot (-2)

    \frac{dy}{dx} = 3 \cdot ln(3x - 1) \cdot 3^{3x - 1} - 2 \cdot ln(1 - 2x) -2

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Amanda-UK View Post
    2) y= 2x-1 -log4(x+1)
    3x+4
    I presume:
    y = \frac{2x - 1}{3x + 4} - log_4(x + 1)

    I, personally, never remember the formula for taking the derivative of a log function. What I do remember is the change of base formula:
    log_b(x) = \frac{log_a(x)}{log_a(b)}
    (This comes in very handy doing logs on my calculator.)

    So I'm going to convert log_4(x) into a ln(x) term, which we know how to take the derivative of.

    log_4(x) = \frac{ln(x)}{ln(4)}

    Thus
    \frac{d}{dx}log_4(x) = \frac{d}{dx} \left ( \frac{ln(x)}{ln(4)} \right ) = \frac{1}{ln(4)} \cdot \frac{1}{x}

    So back to your problem:
    y = \frac{2x - 1}{3x + 4} - log_4(x + 1)

    \frac{dy}{dx} = \frac{(2)(3x + 4) - (2x - 1)(3)}{(3x + 4)^2} - \frac{1}{ln(4)} \cdot \frac{1}{x + 1} \cdot 1

    \frac{dy}{dx} = \frac{11}{(3x + 4)^2} - \frac{1}{ln(4)} \cdot \frac{1}{x + 1}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Amanda-UK View Post
    Find dy/dx of the following:
    3)
    y= e^x (2x+1)^1/2
    Note that \frac{d}{dx}e^x = e^x.

    Thus by the product rule:
    \frac{dy}{dx} = e^x \cdot (2x + 1)^{1/2} + e^x \cdot \frac{1}{2} \cdot (2x + 1)^{-1/2} \cdot (2)

    \frac{dy}{dx} = e^x \cdot (2x + 1)^{1/2} + e^x \cdot \frac{1}{(2x + 1)^{1/2}}

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by Amanda-UK View Post
    Find dy/dx of the following:
    4) intergrate and differentiate:
    xIn(x+1)
    First the derivative:
    \frac{d}{dx} (x \cdot ln(x + 1) ) = (1) \cdot ln(x + 1) + x \cdot \frac{1}{x} = ln(x + 1) + 1

    Now the integral. I'm going to do this using integration by parts:
    \int u dv = uv - \int v du

    So
    \int x ln(x + 1) dx :
    I'm going to set
    u = ln(x + 1) \implies du = \frac{dx}{x + 1}
    dv = x dx \implies v = \frac{1}{2}x^2

    Thus
    \int x ln(x + 1) dx = \frac{1}{2}x^2 \cdot ln(x + 1) - \int \left ( \frac{1}{2}x^2 \right ) \left ( \frac{dx}{x + 1} \right )

    Which leaves us with the \int \frac{x^2}{x + 1} dx, which you can do with partial fractions, or any other method of your choosing.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Aug 2007
    Posts
    22
    thank you as helped me a great deal
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Quote Originally Posted by topsquark View Post
    \int \frac{x^2}{x + 1} dx
    This is the simplest method

    \int\frac{x^2}{x+1}\,dx=\int\frac{x^2-1}{x+1}\,dx+\int\frac1{x+1}\,dx=\int(x-1)\,dx+\int\frac1{x+1}\,dx

    Regards.

    K.
    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