Results 1 to 4 of 4

Thread: d/dtheta

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    11

    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)+ ..... ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Sep 2009
    Posts
    40
    Thanks
    5
    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 $\displaystyle \theta$ is a constant, we are not going to explicitly differentiate it, since $\displaystyle dk/dx$, where k is a constant, is 0. For the first term $\displaystyle 9sec(\theta)$, remember that $\displaystyle d/dx * sec(x) = sec(x)tan(x)$.

    For the second term, remember that $\displaystyle d/dx e^x = dx e^x$. So if $\displaystyle x = \theta cot(\theta)$, then $\displaystyle dx = \theta * -csc^2(\theta)$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor harish21's Avatar
    Joined
    Feb 2010
    From
    Dirty South
    Posts
    1,036
    Thanks
    10
    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)+ ..... ?
    $\displaystyle \frac{d}{d\theta}9e^{\theta}cot{\theta} =
    9[cot{\theta}\frac{d}{d\theta}e^{\theta} + e^{\theta}\frac{d}{d\theta}cot{\theta}]$

    $\displaystyle = 9 [ e^{\theta}cot{\theta} - e^{\theta} csc^2{\theta} ] $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,781
    Thanks
    3030
    Quote Originally Posted by macosxnerd101 View Post
    Since $\displaystyle \theta$ is a constant, we are not going to explicitly differentiate it, since $\displaystyle dk/dx$, where k is a constant, is 0. For the first term $\displaystyle 9sec(\theta)$, remember that $\displaystyle d/dx * sec(x) = sec(x)tan(x)$.

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

Similar Math Help Forum Discussions

  1. dtheta or dx?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 22nd 2010, 06:18 PM

Search Tags


/mathhelpforum @mathhelpforum