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Thread: Elementary Differentiation

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

    Elementary Differentiation

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

    1. Find and simplify the derivative.

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

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

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

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

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

    The answers I've got for these are:

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

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

    2) $\displaystyle y=0.866x-3.128$

    3) $\displaystyle \frac{17}{3}$

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

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

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

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

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

    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 $\displaystyle (-2)^3-\frac{4}{3}(-2)= -8+\frac{8}{3}= -\frac{16}{3}$ and $\displaystyle 3^3-\frac{4}{3}(3)=27-4=23$

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

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

    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
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