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Logarithmic Differentiation Help NeededA LOT OF PROCESS WORK IS SHOWN! please help!

  1. Mar 9, 2009 #1
    Use logarithmic differentiation to find:

    a.) d/dx of [(sin^-1(x^2)*sinh^-1(x^2))/(sin^4(x^2))]

    b.) d^2/dx^2 (sech^-1(e^(2*x)))


    work shown for a:

    let y= [(sin^-1(x^2)*sinh^-1(x^2))/(sin^4(x^2))]

    taking the natural logarithm of both sides:

    ln y= ln [(sin^-1(x^2)*sinh^-1(x^2))/(sin^4(x^2))]

    ln y= ln (sin^-1(x^2)*sinh^-1(x^2)) - ln (sin^4(x^2))

    ln y = ln (sin^-1(x^2)) +ln (sinh^-1(x^2)) - 4*ln(sin^4(x^2))

    differentiating both sides:

    1/y* y'= (1/sin^-1(x^2))(1/sqrt(1-(x^2)^2)(2x) + (1/sinh^-1(x^2))(2x)(1/sqrt(1+(x^2)^2)) - (4)(1/sin(x^2))(2x*cos(x^2))

    substituting y i got:

    y'= (sin^4(x^2)*2x)/(sqrt(1-x^4)) + 2xsin^4(x^2)/(sqrt(1+x^4)) - 8x*cos(x^2)sin^3(x^2)

    is this correct... i think I may have made a few errors

    work shown for b:

    let y= sech^-1(e^(2x))

    taking natural logarithm of both sides:

    ln y= ln (sech^-1(e^(2x))

    differentiating both sides:

    y'= 2/(sqrt(1-e^(4x))

    to find d^2/dx^2 i set dy/dx as y again:

    therefore let y= 2/(sqrt(1-e^(4x))

    finding the natural logarithm of both sides of dy/dx

    i get ln y= ln 2- (1/2)ln (1-e^(4x))

    differentiating both sides leads to:

    y"= 4e^(4x)/((1-e^(4x))^(3/2))

    is this correct.. i believe i may have made a few errors.. please check

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 10, 2009 #2
    Re: Logarithmic Differentiation Help Needed..A LOT OF PROCESS WORK IS SHOWN! please h

    I looked over a and I think that it is pretty good, but I think you made errors when you plugged y back in.
     
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