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Limits and Imp/Explicit Differentiation

  1. Oct 26, 2005 #1
    *opps fixing the latex code*
    I have a test coming up so I tried solving some review questions but couldn't solve all of them. I will also post the ones which I am not sure whether they are correct or wrong so please spare some time to check them too. :smile:
    1.
    [tex] F(x)=
    \left{
    \begin{array}{cc}
    -x, & \mbox{ if } x<-1\\
    x^2, & \mbox{ if } -1 \leq x<1\\
    \frac{1}{x}, & \mbox{ if } x\geq 1
    \end{array}
    \right
    [/tex]
    Use limit definition of the derivative to see if the fuction is differentiable at [tex]x=-1[/tex] and [tex]x=1[/tex]
    At [tex]x=-1[/tex]
    [tex]
    \lim_{x\rightarrow -1^-} \frac{f(x)-f(-1)}{x-(-1)}=-1[/tex]
    [tex]\lim_{x\rightarrow -1^+} \frac{f(x)+1}{x+1} = \frac{(x+1)(x-1)}{x+1}=x-1=-2[/tex]
    If this solution for [tex]x=-1[/tex] is correct then it is also correct for [tex]x=1[/tex] so I won't show my solution for that.
    So the fuction is not differentiable at both points?
    2. Probably correct.
    3. [tex]y=(t^{2/3}).e^{sin(t)}+t[/tex]
    Find the slope at (0,0)
    [tex]\frac{dy}{dx}=\left(\frac{2}{3}t^{-1/3}e^{sin(t)}+t^{2/3}e^{sin(t)}cos(t)\right)+1[/tex]
    [tex]\frac{dy}{dx}=1?[/tex]

    More will come. :)
     
    Last edited: Oct 26, 2005
  2. jcsd
  3. Oct 26, 2005 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, both of those are correct.
     
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