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Math Help - d/dtheta

  1. #1
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    d/dtheta

    h(theta)= 9sec(theta)+9e^(theta)cot(theta)

    h'(theta)=d/dtheta(9sec(theta)) + d/dtheta(9e^(theta)cot(theta))

    ?= 9sec(theta) 9 tan(theta)+ ..... ?
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  2. #2
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    Quote Originally Posted by vjelmy View Post
    h(theta)= 9sec(theta)+9e^(theta)cot(theta)

    h'(theta)=d/dtheta(9sec(theta)) + d/dtheta(9e^(theta)cot(theta))

    ?= 9sec(theta) 9 tan(theta)+ ..... ?
    Since \theta is a constant, we are not going to explicitly differentiate it, since dk/dx, where k is a constant, is 0. For the first term 9sec(\theta), remember that d/dx * sec(x) = sec(x)tan(x).

    For the second term, remember that d/dx e^x = dx e^x. So if x = \theta cot(\theta), then dx = \theta * -csc^2(\theta).
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by vjelmy View Post
    h(theta)= 9sec(theta)+9e^(theta)cot(theta)

    h'(theta)=d/dtheta(9sec(theta)) + d/dtheta(9e^(theta)cot(theta))

    ?= 9sec(theta) 9 tan(theta)+ ..... ?
    \frac{d}{d\theta}9e^{\theta}cot{\theta} =<br />
9[cot{\theta}\frac{d}{d\theta}e^{\theta} + e^{\theta}\frac{d}{d\theta}cot{\theta}]

     = 9 [ e^{\theta}cot{\theta} - e^{\theta} csc^2{\theta} ]
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  4. #4
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    Quote Originally Posted by macosxnerd101 View Post
    Since \theta is a constant, we are not going to explicitly differentiate it, since dk/dx, where k is a constant, is 0. For the first term 9sec(\theta), remember that d/dx * sec(x) = sec(x)tan(x).

    For the second term, remember that d/dx e^x = dx e^x. So if x = \theta cot(\theta), then dx = \theta * -csc^2(\theta).
    Where did you get the idea that \theta was a constant? \theta is the only variable in the problem! And what in the world does the derivative of e^x have to do with this?
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