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Calculating derivative

  1. May 23, 2015 #1
    1. The problem statement, all variables and given/known data
    [tex]f(x,y)=cos(xy)+ye^{x}[/tex] near [tex](0,1)[/tex], the level curve [tex]f(x,y)=f(0,1)[/tex] can be described as [tex]y=g(x)[/tex], calculate [tex]g'(0)[/tex].

    2. Relevant equations
    N/A
    Answer is -1.
    3. The attempt at a solution
    If you do [tex]f(0,1)=cos((0)(1))+1=2[/tex], do you have to use linear approximation or some other method?
     
  2. jcsd
  3. May 23, 2015 #2

    HallsofIvy

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    Yes, at f(0, 1)= 2 so the level curve is [itex]cos(xy)+ e^xy= 2[/itex]. Now, what they are calling "g(x)" is y(x) implicitly defined by that (that is, you could, theoretically, solve that equation for y.) Use implicit differentiation to find y'(x) from that.
     
  4. May 23, 2015 #3
    Why don't you first find an expression for ##\frac{d(F(x,y))}{dx}## ? You don't have to use any approximation methods here.
     
  5. May 23, 2015 #4
    So implicitly differentiate [tex]cos(xy)+e^{x}y=2[/tex]?

    I got something along the lines of [tex]\frac{dy}{dx}=\frac{ysin(xy)-ye^{x}}{-xsin(xy)+e^{x}}[/tex].
     
  6. May 23, 2015 #5
    That is correct. Now substitute the ##x## and ##y## values and you'll get your answer. (Remember that for ##f(0,1)=2## , ##g(x)=y## . You've just found an expression for ##g'(x)##.)
     
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