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Math Help - Logarithmic Differentiation

  1. #1
    Newbie
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    May 2009
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    Gauteng,Johannesburg
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    Logarithmic Differentiation

    I came across this problem :

    y = 7x (cosx)^x/2 ;

    where I'm to find y'. So with the understading that (cosx) is raised to a variable power of x/2, I get an idea of applying Logarithmic Differentiation as follows :

    Iny = In [7x (cosx)^x/2],

    then apply the Log Law saying that the Log of a Product is the Log of the Sum to have :

    Iny = In (7x) + x/2 In(cosx), then go on to finding dy/dx by finding the derivative of the sum,in which I apply the Product Rule to find dy/dx of x/2 In(cosx) to get :

    dy/dx Iny = dy/dx[In 7x] + [In(cosx)dy/dx (x/2) + (x/2) dy/dx In(cosx)]

    => dy/dx Iny=(1/7x)7 + [In(cosx)/2 - x(sinx)/2(cosx)]
    => dy/dx Iny=(1/x) + 1/2[In(cosx) - x(tanx)]
    => dy/dx = y {(1/x) + 1/2[In(cosx) - x(tanx)]},
    Then by substituting the given y value, I obtain :

    dy/dx = [7x(cosx)^x/2]{(1/x) + 1/2[In(cosx) - x(tanx)]}:];
    Which I regard as my solution. Do I need to simplify any further ?
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  2. #2
    Senior Member
    Joined
    Jan 2010
    Posts
    354
    Your work looks correct.

    Whether or not you are required to simplify your solution would usually be determined by your teacher. But if you are asking whether it can be simplified any further, I don't see anything that would reduce.

    I will point out though that the correct symbol is "ln" (lowercase L, not capital i).
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