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Math Help - Find the derivative of the function y = tan...

  1. #1
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    Find the derivative of the function y = tan...

    Find the derivative of the function.

    y = tan(9θ)^2

    y'(θ) = ?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by question111 View Post
    Find the derivative of the function.

    y = tan(9θ)^2

    y'(θ) = ?
    by the chain rule: \frac d{dx} \tan u = u' \sec^2 u where u is a function of x

    does that help?
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  3. #3
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    Mmm, not exactly. I'm quite behind in my work, and I don't really see how to use the info you gave me there...

    I'm particularly confused by the symbol θ ?
    I honestly don't know what that symbol is supposed to mean for this equation.
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  4. #4
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    the symbol looks unclear in the problem but is it suppose to look like a zero with a horizontal line in it? if so its just a variable. it is like having x in there.
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  5. #5
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    Quote Originally Posted by question111 View Post
    Find the derivative of the function.

    y = tan(9θ)^2

    y'(θ) = ?
    y = \tan^2{\theta} = (\tan{\theta})^2.

    Let u = \tan{\theta}, then y = u^2.


    Now use the chain rule, \frac{dy}{d\theta} = \frac{du}{d\theta}\times \frac{dy}{du}.

    \frac{du}{d\theta} = \sec^2{\theta}

    \frac{dy}{du} = 2u = 2\tan{\theta}

    So \frac{dy}{d\theta} = 2\tan{\theta}\sec^2{\theta}.


    Hope that helped.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Prove It View Post
    y = \tan^2{\theta} = (\tan{\theta})^2.

    Let u = \tan{\theta}, then y = u^2.


    Now use the chain rule, \frac{dy}{d\theta} = \frac{du}{d\theta}\times \frac{dy}{du}.

    \frac{du}{d\theta} = \sec^2{\theta}

    \frac{dy}{du} = 2u = 2\tan{\theta}

    So \frac{dy}{d\theta} = 2\tan{\theta}\sec^2{\theta}.


    Hope that helped.
    the OP seemed to write y = \tan (9 \theta)^2 as opposed to \tan^2 9 \theta. i wonder if it was a typo....?
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