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Homework Help: Differentiation questions

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Find dy/dx for:

    a) xsin(xy^2) - ln(x/y) = y

    b) x = 3cos(a), y = 2sec(a). simplify then find dy/dx when a = pi/3



    2. Relevant equations



    3. The attempt at a solution

    I've done both of them, was just hoping someone could check I've done things correctly. I'm still feeling pretty uncertain.


    a)

    xsin(xy^2) - ln(x/y) = y

    dy/dx = d/dx[xsin(xy^2) - ln(x/y)]

    dy/dx = sin(xy^2) + (xcos(xy^2) * (y^2 + 2xy*dy/dx)) - 1/(x/y) * (1/y - x/(y^2) *dy/dx)

    dy/dx = sin(xy^2) + xy^2cos(xy^2) + 2x^2ycos(xy^2)dy/dx - (1/y)/(x/y) + (x/y^2)/(x/y) * dy/dx

    thus

    dy/dx = (-sin(xy^2) - xy^2cos(xy^2) +1/x) / (2x^ycos(xy^2) + 1/y - 1)



    b)

    dy/dx = (dy/da) / (dx/da)

    = 2sec(a)tan(a) / -3sin(a)

    using tan(a) = sin(a) / cos(a), and sec(a) = 1/cos(a)

    dy/dx = -(2/3) * 1/cos(a) * sin(a)/cos(a) * 1/sin(a)

    = -(2/3) * (1/cos^2(a))

    so for a = pi/3

    dy/dx = -(2/3) * (1/ (.5^2))
    = -(2/3) * 4
    = -2 2/3
     
  2. jcsd
  3. Oct 12, 2008 #2

    arildno

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    There might be some sign mistakes lurking in there that I didn't catch, but your general procedure is sound.

    At b), why not just use:
    [tex]\cos(a)=\frac{x}{3}\to\sec(a)=\frac{3}{x}\to{y}=\frac{6}{x}\to\frac{dy}{dx}=-\frac{6}{x^{2}}[/tex]
    Then, a=pi/3 implies x=3/2, whereby [tex]\frac{dy}{dx}=-\frac{8}{3}[/tex]
     
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