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Math Help - Implicit Differentiation of a Trig Function?

  1. #1
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    Implicit Differentiation of a Trig Function?

    For some reason, I'm having A LOT of problems differentiating trig functions, whether or not they're implicit.

    sin x + 2 cos 2y = 1

    I got as far as cos x - 2 sin 2y... and then I feel like there's another term added on here.

    I have the answer (from the book), I just don't have the middle steps to understand what happened.
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  2. #2
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    Quote Originally Posted by Ohoneo View Post
    For some reason, I'm having A LOT of problems differentiating trig functions, whether or not they're implicit.

    sin x + 2 cos 2y = 1

    I got as far as cos x - 2 sin 2y... and then I feel like there's another term added on here.

    I have the answer (from the book), I just don't have the middle steps to understand what happened.
    Taking the implicit derivative gives

    \cos(x)-4\sin(2y)\frac{dy}{dx}=0
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    But where does the 4(sin y) come from?

    For example, my book writes this:

    sin x + 2 cos 2y = 1
    cos x - 4(sin 2y)dy/dx = 0
    dy/dx = cos x/4 sin 2y

    I understand how to get from step 2 to step 3, and I understand part of step two (cos x is the derivative of sin x, so that makes sense, and the negative part of the next part comes from the fact that the derivative of cos x is -sin x), but how exactly does the 4(sin 2y) come about?
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    Quote Originally Posted by Ohoneo View Post
    But where does the 4(sin y) come from?

    For example, my book writes this:

    sin x + 2 cos 2y = 1
    cos x - 4(sin 2y)dy/dx = 0
    dy/dx = cos x/4 sin 2y

    I understand how to get from step 2 to step 3, and I understand part of step two (cos x is the derivative of sin x, so that makes sense, and the negative part of the next part comes from the fact that the derivative of cos x is -sin x), but how exactly does the 4(sin 2y) come about?
    You need to use the chain rule don't forget to take the derivative of the inside

    \frac{d}{dx}f(2y)=\frac{df}{dy}(2\frac{dy}{dx})

    \frac{d}{dx}2\cos(2y)=2(-\sin(2y))\frac{d}{dx}(2y)=-4\sin(2y)\frac{dy}{dx}
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    Ah, thank you so much. That's what I was missing. Thank you!
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