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Math Help - Trig Derivative Question

  1. #1
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    Trig Derivative Question

    y'(2\sin^{2}\theta) = ??
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    Super Member Matt Westwood's Avatar
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    Re: Trig Derivative Question

    Quote Originally Posted by Jason76 View Post
    y'(2\sin^{2}\theta) = ??
    What is y differentiated with respect to? Write it out as a dy / d? form, or we can't see what's being derived with respect to what.
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    Re: Trig Derivative Question

    Maybe he meant the derivative of 2sin^2(Θ)? Then it's just 2*2sin(Θ)*cos(Θ) = 2sin(2Θ). Otherwise, I don't know...
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    Re: Trig Derivative Question

    Your notation is very confusing. y'(2 sin(2\theta)) means the derivative of some unknown function, y(x), evaluated at x= sin(2\theta). I believe what you mean is " y(\theta)= 2 sin(2\theta). Find the derivative of y with respect to \theta". To do that, you need to know:
    1) the derivative of 2x with respect to x.
    2) the derivative of sin(x) with respect to x.
    3) the derivative of 2\theta with respect to \theta and
    4) the chain rule.

    Do you know those?
    Last edited by HallsofIvy; July 22nd 2014 at 04:47 AM.
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    Re: Trig Derivative Question

    y = 2\sin^{2}\theta

    y' = 2\(u)^{2} du

    u =  \sin \theta

    du = \cos\theta

    y' = 4\sin\theta\cos\theta
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    Super Member Matt Westwood's Avatar
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    Re: Trig Derivative Question

    Quote Originally Posted by Jason76 View Post
    y = 2\sin^{2}\theta

    y' = 2\(u)^{2} du

    u =  \sin \theta

    du = \cos\theta

    y' = 4\sin\theta\cos\theta
    du = \cos\theta as it stands is meaningless and wrong. So is y' = 2\(u)^{2} du which is even worse.

    You need to write \frac{du}{d \theta} = \cos\theta and y = 2 (u)^{2}.

    Please revise your notion of derivatives and exactly what the notation means. Many of your problems in understanding what's going on may stem from your continued inaccuracy with your notation.

    Steer clear of the y' notation unless it is completely clear what y is derived with respect to. Here it is not clear at all.
    Last edited by Matt Westwood; July 23rd 2014 at 12:02 AM.
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