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More differentiation problems

  1. Aug 30, 2005 #1
    1. If [itex]x = \cos \theta[/itex] and [itex]y = \cos \theta + \sin \theta[/itex], [itex]\frac{dy}{dx}[/itex] is? (Answer: [itex]1 - \cot \theta[/itex])

    I'm just not seeing this question... any help appreciated.

    2. [itex]y = \ln(\sec 2x + \tan 2x)[/itex] (Answer = [itex]2\sec 2x[/itex])

    I get to [tex]\frac{1}{(2\sec x\tan x) + (2\sec^2 x)}[/tex] but it doesn't seem that I'm on the right track compared to the answers I have.
    Last edited: Aug 30, 2005
  2. jcsd
  3. Aug 30, 2005 #2


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    1) If [itex]x = \cos \theta[/itex] then [itex]dx = - \sin \theta[/itex], right?
    Now compute [itex]dy[/itex] as well and do the division :wink:

    2) What do you have to do here? Just compute [itex]dy[/itex] or [itex]dy/dx[/itex] again with the x from 1?
  4. Aug 30, 2005 #3
    Thanks I got 1 now! As for 2, it's a seperate question. I just need to differentiate with respect to x.
  5. Aug 30, 2005 #4
    For your second one, you need to remember that [tex]\frac{d(\ln{x})}{dx}=\frac{1}{x}*dx[/tex] Just remember to use the chain rule and you should get the right answer.
  6. Aug 30, 2005 #5


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    For the derivative of an ln of a function, you can use the following (follows from the chain rule)

    [tex]\left(\ln \left( {f\left( x \right)} \right)\right)^\prime = \frac{{f'\left( x \right)}}
    {{f\left( x \right)}}[/tex]
  7. Aug 30, 2005 #6
    OK, so I should get [tex]\frac{(2 \sec 2x \tan 2x) + (2 \sec^2 2x)}{(\sec 2x + \tan 2x)}[/tex] but I still can't get it to [itex]2 \sec 2x[/tex], my algebra is failing me! :blushing:
    Last edited: Aug 30, 2005
  8. Aug 30, 2005 #7


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    For the derivative

    [tex]\left( {\sec \left( {2x} \right) + \tan \left( {2x} \right)} \right)^\prime = \frac{{2\sin \left( {2x} \right)}}{{\cos ^2 \left( {2x} \right)}} + \frac{2}{{\cos ^2 \left( {2x} \right)}} = \frac{{2\sin \left( {2x} \right) + 2}}{{\cos ^2 \left( {2x} \right)}}[/tex]

    Then it's just putting everything into sines and cosines

    [tex]\frac{{\frac{{2\sin \left( {2x} \right) + 2}}{{\cos ^2 \left( {2x} \right)}}}}{{\sec \left( {2x} \right) + \tan \left( {2x} \right)}} = \frac{{\frac{{2\sin \left( {2x} \right) + 2}}{{\cos ^2 \left( {2x} \right)}}}}{{\frac{1}{{\cos \left( {2x} \right)}} + \frac{{\sin \left( {2x} \right)}}{{\cos \left( {2x} \right)}}}} = \frac{{\frac{{2\left( {1 + \sin \left( {2x} \right)} \right)}}{{\cos ^2 \left( {2x} \right)}}}}{{\frac{{1 + \sin \left( {2x} \right)}}{{\cos \left( {2x} \right)}}}} = \frac{2}{{\cos \left( {2x} \right)}} = 2\sec \left( {2x} \right)[/tex]
  9. Aug 30, 2005 #8
    Wow, I never would have figured it out thinking the way I was. Thank you for being patient with me :smile:
  10. Aug 30, 2005 #9


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    No problem.

    Normally I'd have tried to guide you more through it but I hope my previous posts is clear enough and that you actually understand what happened.

    Since you try to simpplify, it is easier to lose the tangent and secans and switch all over to sines and cosines. Then it's purely algebra, simplifying the expression.
  11. Aug 30, 2005 #10


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    take out [itex]2 \sec 2x[/tex] as a factor from the numerator. The other bits will cancel.
  12. Aug 30, 2005 #11
    Ah, I never thought of that either. Thanks.
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