Results 1 to 2 of 2

Math Help - Elementary Differentiation

  1. #1
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823

    Elementary Differentiation

    (Be sure to use differential notation properly. Show all work for full credit)

    1. Find and simplify the derivative.

    a) f(x)=4\cos{x}+x^{4/3}

    b) y=\frac{1}{x}+5e^x


    2) Find the equation of the line tangent to y=\cos{x} at (\frac{4\pi}{3},-\frac{1}{2}). Give exact coefficients and approximate coefficients to the nearest thousandth.


    3) Find the average rate of change of f(x)=x^3-\frac{4}{3}x on the interval [-2,3].


    4) Find the instantaneous rate of change of w with respect to z if w=\frac{7}{3z^2}.


    The answers I've got for these are:

    1) a) \frac{4}{3}x^{1/3}-4sinx

    b) 5e^x-\frac{1}{x^2}

    2) y=0.866x-3.128

    3) \frac{17}{3}

    4) -\frac{14}{3z^3}

    What I would like for someone to do is:

    1) Tell me if my answers are correct
    2) Do the problems so that I can see how you did them because I got the answers, but it wasn't pretty.
    3) tell me if my answers are simplified they way they should be.

    Thanks Dudes. (and dudesses)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    May 2009
    Posts
    471
    y=4cos(x)+x^{\frac{4}{3}}\rightarrow y'= -4sin(x)+\frac{4}{3}x^{\frac{4}{3}-1}= -4sin(x)+\frac{4}{3}x^{\frac{1}{3}}

    y=\frac{1}{x}+5e^x\rightarrow y'=\frac{x*0-1*1}{x^2}+5e^x=5e^x-\frac{1}{x^2}

    y=cos(x)\rightarrow y'=-sin(x) and this must be the slope of the tangent line at the point (x,y). So we plug in our x-value of \frac{4\pi}{3} and see that y'=-sin{\frac{4\pi}{3}}=\frac{\sqrt{3}}{2} So we now use a well known formula for a line, point slope form

    y-(-\frac{1}{2})=\frac{\sqrt{3}}{2}(x-\frac{4\pi}{3})\rightarrow y+\frac{1}{2}=\frac{\sqrt{3}}{2}(x-\frac{4\pi}{3})

    The average rate of change is just the calculation of the slope between the end point. So x=-2 and x=3 are the x coordinated of the endpoints, and the y values are (-2)^3-\frac{4}{3}(-2)= -8+\frac{8}{3}= -\frac{16}{3} and 3^3-\frac{4}{3}(3)=27-4=23

    And now plug into the slope formula to get \frac{23-\left(-\frac{16}{3}\right)}{3-(-2)}=\frac{17}{3}

    w=\frac{7}{3z^2}\rightarrow w'=\frac{3z^2*0-7(6z)}{(3z^2)^2}=-\frac{42z}{9z^4}=\frac{14}{3z^3}

    So yeah... you're answers are good, everything looks simple, and there's all the work you would need to show for full credit, you could prolly even skip some steps in there, but as this is a show me how question, I don't like to skip steps
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Elementary Row Operations
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: October 9th 2010, 03:44 AM
  2. The product log is not elementary
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 24th 2010, 06:51 PM
  3. Elementary Row Operations
    Posted in the Advanced Algebra Forum
    Replies: 13
    Last Post: June 10th 2008, 10:34 AM
  4. Elementary Analysis
    Posted in the Calculus Forum
    Replies: 9
    Last Post: February 20th 2007, 11:47 AM
  5. Is this elementary.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 19th 2006, 06:00 PM

Search Tags


/mathhelpforum @mathhelpforum