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Find equation tangent to y=(x^2)lnx at (1,0)

  1. Jan 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Find an equation for the line y=(x^2)lnx at (1,0)


    2. Relevant equations



    3. The attempt at a solution
    I took the first derivative of the equation and got y'=2xlnx +x
    In other equations, it would be simple to find the slope, but at this point I am lost, is the slope 2? It doesn't appear to be in the form of y=mx+b.

    So I've decided to use f'(1)=2(1)ln1 + 1 = 1 , this gives me the slope of the function at the point (1,f(1)) right?

    If that's the case, then
    y=m(x-x_1) + y_1
    = 1(x-1) + 0
    y=x-1

    Just wanted to know if I am close or way off or maybe correct.
     
  2. jcsd
  3. Jan 26, 2014 #2

    Dick

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    Science Advisor
    Homework Helper

    Looks ok to me.
     
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