Results 1 to 7 of 7

Math Help - [SOLVED] Differentiation

  1. #1
    Member Awsom Guy's Avatar
    Joined
    Jun 2009
    From
    Mars
    Posts
    186

    [SOLVED] Differentiation

    I have been trying to do this question for like a week, and I have had no success.
    Question:
    Differentiate the function f(x)=(4x)/(sqrt2x)
    I have tried it and I know that I have to use the quotient rule but I know I am doing something wrong in this step.

    (2x)^1/2 * (4) - (4x) * (x)^-1/2 all divided by (2x)^3/2

    Please help me do the multypling out.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    Quote Originally Posted by Awsom Guy View Post
    I have been trying to do this question for like a week, and I have had no success.
    Question:
    Differentiate the function f(x)=(4x)/(sqrt2x)
    I have tried it and I know that I have to use the quotient rule but I know I am doing something wrong in this step.

    (2x)^1/2 * (4) - (4x) * (x)^-1/2 all divided by (2x)^3/2

    Please help me do the multypling out.
    f(x) = \frac{4x}{\sqrt{2x}}

     = 4x(2x)^{-\frac{1}{2}}.

    Now using the product rule:

    f'(x) = 4x\frac{d}{dx}[(2x)^{-\frac{1}{2}}] + (2x)^{-\frac{1}{2}}\frac{d}{dx}(4x)

     = 4x\cdot 2\cdot \left(-\frac{1}{2}(2x)^{-\frac{3}{2}}\right) + 4(2x)^{-\frac{1}{2}}

     = 4(2x)^{-\frac{1}{2}} - 4x(2x)^{-\frac{3}{2}}

     = \frac{4}{\sqrt{2x}} - \frac{4x}{(\sqrt{2x})^3}

     = \frac{4}{\sqrt{2x}}\left[1 - \frac{x}{(\sqrt{2x})^2}\right]

     = \frac{4}{\sqrt{2x}}\left(1 - \frac{x}{2x}\right)

     = \frac{4}{\sqrt{2x}}\left(1 - \frac{1}{2}\right)

     = \frac{2}{\sqrt{2x}}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Awsom Guy's Avatar
    Joined
    Jun 2009
    From
    Mars
    Posts
    186

    helps

    Quote Originally Posted by Prove It View Post
    f(x) = \frac{4x}{\sqrt{2x}}

     = 4x(2x)^{-\frac{1}{2}}.

    Now using the product rule:

    f'(x) = 4x\frac{d}{dx}[(2x)^{-\frac{1}{2}}] + (2x)^{-\frac{1}{2}}\frac{d}{dx}(4x)

     = 4x\cdot 2\cdot \left(-\frac{1}{2}(2x)^{-\frac{3}{2}}\right) + 4(2x)^{-\frac{1}{2}}

     = 4(2x)^{-\frac{1}{2}} - 4x(2x)^{-\frac{3}{2}}

     = \frac{4}{\sqrt{2x}} - \frac{4x}{(\sqrt{2x})^3}

     = \frac{4}{\sqrt{2x}}\left[1 - \frac{x}{(\sqrt{2x})^2}\right]ummm.... how did the power go from 3 to 2.

     = \frac{4}{\sqrt{2x}}\left(1 - \frac{x}{2x}\right)

     = \frac{4}{\sqrt{2x}}\left(1 - \frac{1}{2}\right)

     = \frac{2}{\sqrt{2x}}.
    PLease explain.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    Quote Originally Posted by Awsom Guy View Post
    PLease explain.
    I took out \frac{4}{\sqrt{2x}} as a common factor.

    Note that (\sqrt{2x})^3 = \sqrt{2x}(\sqrt{2x})^2.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Awsom Guy's Avatar
    Joined
    Jun 2009
    From
    Mars
    Posts
    186
    Oh ok thanks for that. Note: I have been stuck with that question for almost a week. Thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,389
    Thanks
    1324
    Like Prove It, I recommend using negative powers and the product rule rather than positive powers and the quotient rule. However, since you specifically asked about that:

    (2x)^1/2 * (4) - (4x) * (x)^-1/2 all divided by (2x)^3/2

    \frac{(2x)^{1/2}(4)- (4x)(x)^{-1/2}}{(2x)^{3/2}}

    You have a typo in the second term of the numerator: it should be [tex](2x)^{-1/2}[tex], not (x)^{-1/2}. Also the denominator is wrong. In the quotient rule, the quotient of the derivative is the square of the denominator of the function. Here, that is ((2x)^{1/2})^2= 2x. You should have
    \frac{(2x)^{1/2}(4)- (4x)(2x)^{-1/2}}{2x}


    To simplify that, factor (2x)^{-1/2} out of the numerator.

    (2x)^{1/2}= \frac{(2x)}{(2x)^{-1/2}} so your numerator becomes (2x)^{-1/2}[(2x)(4)- 4x]= (2x)^{-1/2}(4x).

    Putting that into half power into the denominator makes it (2x)^{1/2} and, of course, (2x)^{1/2}(2x)= (2x)^{3/2}.

    You have a 3/2 power in the denominator after simplifying:
    \frac{4x}{(2x)^{3/2}}= \frac{4x}{\left(\sqrt{2x}\right)^3}

    (Since the original problem was given in terms of the square root rather than a fractional power, it is probably better to give the answer that way.)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member Awsom Guy's Avatar
    Joined
    Jun 2009
    From
    Mars
    Posts
    186
    umm..lols thanks I have already solved this :0 thanks anyways Ill close this thread
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] differentiation...
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 10th 2009, 02:08 AM
  2. [SOLVED] Differentiation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 16th 2009, 08:48 AM
  3. [SOLVED] Differentiation.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 20th 2009, 10:34 PM
  4. [SOLVED] Differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 5th 2009, 03:03 PM
  5. [SOLVED] Differentiation Help!
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 29th 2008, 01:50 PM

Search Tags


/mathhelpforum @mathhelpforum