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logarithmic differentiation question

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  • 29-04-2012 10:37pm
    #1
    joeperry
    Closed Accounts Posts: 527 ✭✭✭


    Hi im having trouble with one of these. My notes seem to go wrong.

    y=(2-3x)^sin(x)

    So i get,

    ln(y)=ln(2-3x)^sin(x)

    then,

    ln(y)=sin(x).ln(2-3x)

    Unsure of the next step, do i Substitute u for 2-3x by chain rule? if i do i get 3/2-3x. I can't find any similar examples with powers,thanks!


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    #2
    MathsManiac
    Registered Users Posts: 1,595 ✭✭✭MathsManiac


    You differentiate both sides.

    LHS gives (1/y)(dy/dx).

    For the RHS, use the product rule with u=sin(x) and v=ln(2-3x).


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