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Derivative help#2

  1. Apr 19, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]y= \frac{sin^2 x}{1+ ctgx} +\frac{cos^2 x}{1+ ctgx}[/tex]

    Find: [tex]y'[/tex]

    3. The attempt at a solution

    Answer is[tex]-cos2x[/tex] but again I can't find it.

    Where am I making a mistake?

    [tex] 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} =[/tex]

    = \frac{sin2x( 1+ctgx) +1 }{(1+ ctgx)^2} +\frac{-sin2x(1+tgx)-1}{(1+ tgx)^2}
  2. jcsd
  3. Apr 19, 2009 #2


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    Science Advisor
    Homework Helper

    Well the denominator in the second term goes from being 1 + ctgx to being 1 + tgx.
  4. Apr 19, 2009 #3


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    Homework Helper

    The only mistake you've made so far is the function y you gave us. The denominator of the second term should have a tangent, not a cotangent. You have done this correctly during your differentiation. I am afraid it's just a matter of simplification now by using trig identities.
    Last edited: Apr 19, 2009
  5. Apr 19, 2009 #4


    Staff: Mentor

    Concerning the problem you posted (which is apparently not the right problem), for future reference you could have made life easier by doing some simplification before doing the differentiation.
    [tex]y= \frac{sin^2 x}{1+ ctgx} +\frac{cos^2 x}{1+ ctgx}[/tex]
    [tex]\Leftrightarrow y = \frac{sin^2 x + cos^2 x}{1+ ctgx}[/tex]
    [tex]\Leftrightarrow y = \frac{1}{1+ ctgx}[/tex]
    Now you have a much simpler expression to differentiate, which makes it less likely that you'll make errors.
  6. Apr 19, 2009 #5
    Yes, it's my mistake. Denominator of the second term is tangent.
  7. Apr 19, 2009 #6


    Staff: Mentor

    Right, I got that. The point of my post was that you made the (incorrect) problem harder than you needed to by not simplifying first. Always try to do things the simplest way you can get away with.
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