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Calculus II - closed function question

  1. Jan 31, 2012 #1
    there is a function: F ( x, y, z) = 2ln (xz) + sin ( xyz) − y^2 = 0.
    the func is defined by the closed function z=f(x,y) and provides : f(1,0)=1
    we define: g(t)=f(t,1-t^6) . where t is very close to 1.
    I have to find g'(1)

    2. Relevant equations

    I tried to to do like that: find F'x and F'z and did z'x =-(F'x/ F'z) and got -1. but from here I dont know what to do.
    The answer is g'(1)=2.
    Thanks for your help!
    Last edited: Jan 31, 2012
  2. jcsd
  3. Jan 31, 2012 #2


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    Homework Helper

    sorry, can't really follow this - can you explain the question in a little more detail and more clearly? for example what is g in relation to F and f and x in relation to t?
  4. Jan 31, 2012 #3
    sorry about the wrong copying: g(t)=f(t,1-t^6)
  5. Jan 31, 2012 #4


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    Science Advisor

    Are you saying there exist f(x,y) such that z= f(x,y), in the neighborhood of (1, 0, 1), is the same as [itex]2ln(xz)+ sin(xyz)- y^2= 0[/itex]?

    If so, then F(x, y, g)= 2ln(xg)+ sin(xyg)- y^2= 0. And with x= t, y= 1- t^6, that is 2ln(tg)+ sin(t(1- t^6)g)- (1- t^6)^2.

    Differentiating with respect to t,
    [tex]\frac{2}{tg}(1+ tg')+ cos(t(1- t^6)g)((1- t^6)g- 6t^6g+ t(1-t^6)g')- 12(1- t^6)t^5= 0[/tex]
    Set t= 1, so that x= 1, y= 0, and F(1, 0, g(1))= 2ln(g(1))= 0. What is g(1)? Put x= 1, y= 0, and that value of g(1) into the equation and solve for g'(1).

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