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Homework Help: How to find differencial by using implicit functions?

  1. Mar 18, 2013 #1
    1. The problem statement, all variables and given/known data
    find partial equations respect to c. and respect to h
    use implicit function differentiation of the reciprocal of R to answer
    what is the differential change in R when c=20 h=30 and c changes to 21

    2. Relevant equations
    is there any way to make R easier?
    i said that R=ch/(.55h+.45c) which was the best i could do.
    Is there any other way to make R easier?

    3. The attempt at a solution
    i got the partial equations if the R=ch/(.55h+.45c) is right.
    but im not sure how to use implicit or what implicit is. i just found it normally.
    WHen it says what is the differential change in R when c=20 and h=3 and c changes to 21. do i just substitude them to the partial differentials and add them?
  2. jcsd
  3. Mar 18, 2013 #2


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    Yes, R= ch/(.55h+ .45c) is correct and about as simple as it gets. I notice that the problem asks you to "use implicit function differentiation of the reciprocal of R. That is, of course, 0.55h+ 0.45c= chR. Differentiate both sides of that with respect to h to find [itex]\partial R/\partial h[/itex]. (Surely you remember "implicit differentiation" from Calculus I?)

    In general the "differential" of a function, f(h,c), is
    [tex]df= \frac{\partial f}{\partial h}dh+ \frac{\partial f}{\partial c}dc[/tex]

    But notice that, in this problem, only c changes.
  4. Mar 18, 2013 #3
    thank you.
    just one more thing.
    could you walk me through implicit differentiation on this problem?
    i start off when i find dR/dh. i got .55+0=crDr/dh ? is this right or am i missing something?
  5. Mar 18, 2013 #4
    Hello munkhuu, I believe this link would be tremendously helpful in solving your problem: http://tutorial.math.lamar.edu/Classes/CalcIII/ChainRule.aspx

    To use implicit differentiation, you need to first present the given equation in the form F(R,c,h) by isolating the three variables into one side. So in our case, F(R,c,h)=R-[itex]\frac{ch}{0.55h+0.45c}[/itex]=0. And according to the Implicit Differentiation Rule, ∂R/∂c=-(∂F/∂c)/(∂F/∂h) (notice the negative sign!!), where ∂F/∂c=[h*(0.55h+0.45c)-ch*(0.45)]/[0.55h+0.45c]^2 (using the quotient rule and considering h and R constants, we have differentiated F with respect to c); and ∂F/∂R=1. Proceed in a similar fashion and we will get ∂R/∂h.

    As for part b), in order to find the marginal effect of c on R (notice that h remains unchanged), we just need to multiply ∂R/∂c with ∂c to get rid of the denominator, and plug in (c=21, h-30, ∂c (change in c)=1). I hope my approach is right. Good luck :)
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