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Math Help - Derivative Calculation

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    37

    Derivative Calculation

    Given:

       y= \frac{sin^2 x}{1+ ctgx}  + \frac{cos^2 x}{1+ ctgx}

    Find  y'

    I know that answer is   -cos2x , but can't calculate it.

    I got so far:

     y'=(\frac{sin^2 x}{1+ ctgx} +\frac{cos^2 x}{1+ ctgx} )'= \frac{(sin^2 x)'(1+ ctgx) - sin^2 x (1+ ctgx)'}{(1+ ctgx)^2}                   +   \frac{(cos^2 x)'(1+ tgx) - cos^2 x (1+ tgx)'}{(1+ tgx)^2} =

    <br />
=\frac{2sinx(1+ctgx)+1}{(1+ ctgx)^2} +\frac{2cos x(1+tgx)-1}{(1+ tgx)^2}<br />
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  2. #2
    MHF Contributor
    Joined
    Oct 2005
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    I spot one crucial error. There might be more but you need to correct this one first.

    (\cos^2(x))' = -2\cos(x)\sin(x) and (\sin^2(x))' = 2\sin(x)\cos(x)
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