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Math Help - Chain Rule W/Trig

  1. #1
    Member VitaX's Avatar
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    Chain Rule W/Trig

    Find the derivative of y = \left(-1 - \frac{csc \Theta}{2} - \frac{\Theta^2}{4}\right)^2

    So I'm guessing here you just use the chain rule to find the derivative. I get:
    y' = 2 \left(-1 - \frac{1}{2} csc \Theta - \frac{1}{4} \Theta^2\right)\left(\frac{1}{2}csc \Theta cot\Theta - \frac{1}{2}\Theta\right)

    If I know I did this much correct so far I can go on and simplify it.
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  2. #2
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by VitaX View Post
    Find the derivative of y = \left(-1 - \frac{csc \Theta}{2} - \frac{\Theta^2}{4}\right)^2

    So I'm guessing here you just use the chain rule to find the derivative. I get:
    y' = 2 \left(-1 - \frac{1}{2} csc \Theta - \frac{1}{4} \Theta^2\right)\left(\frac{1}{2}csc \Theta cot\Theta - \frac{1}{2}\Theta\right)

    If I know I did this much correct so far I can go on and simplify it.
    Looks okay to me.
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  3. #3
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    Quote Originally Posted by VitaX View Post
    Find the derivative of y = \left(-1 - \frac{csc \Theta}{2} - \frac{\Theta^2}{4}\right)^2

    So I'm guessing here you just use the chain rule to find the derivative. I get:
    y' = 2 \left(-1 - \frac{1}{2} csc \Theta - \frac{1}{4} \Theta^2\right)\left(\frac{1}{2}csc \Theta cot\Theta - \frac{1}{2}\Theta\right)

    If I know I did this much correct so far I can go on and simplify it.


    Well, csc x = 1/sinx, so (csc x)' = -cos x/sin^2(x). Do you really think this is nastier than -csc x*cot x??? Of course, you save yourself a quotient...

    Anyway, the derivation looks fine to me.

    Tonio
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  4. #4
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by tonio View Post
    Well, csc x = 1/sinx, so (csc x)' = -cos x/sin^2(x). Do you really think this is nastier than -csc x*cot x??? Of course, you save yourself a quotient...

    Anyway, the derivation looks fine to me.

    Tonio
    The derivative of \csc x is quoted in all the reference works and textbooks that I've ever come across as -\csc x \cot x. IMO it's perfectly appropriate, and I actually find it easier to learn like that. As you say, it saves a quotient.

    To take it to its logical conclusion, there's no point in learning it anyway, because you can always use the quotient rule on \cos x / \sin x.
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