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cannotdoosmath Group Title

Find the derivative: y= sin(cos(2x))

  • one year ago
  • one year ago

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  1. AccessDenied Group Title
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    Are you familiar with the Chain Rule for differentiation?

    • one year ago
  2. cannotdoosmath Group Title
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    Yes

    • one year ago
  3. cannotdoosmath Group Title
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    Help me along though please

    • one year ago
  4. cannotdoosmath Group Title
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    so sin(2x) * d/dx (sin(2x)?

    • one year ago
  5. AccessDenied Group Title
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    What is the general set up for the Chain Rule? y = f(g(x)) What is the derivative of this general function?

    • one year ago
  6. cannotdoosmath Group Title
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    derivative is cos

    • one year ago
  7. AccessDenied Group Title
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    Hmm... I think you have something mixed up in your approach. The Chain Rule states: y = f(g(x)); the derivative y' = f'(g(x)) * g'(x) In our case (y = sin(cos(2x)))), f(x) = sin x and g(x) = cos (2x). y' = derivative of f with respect to g, times the derivative of g with respect to x.

    • one year ago
  8. AccessDenied Group Title
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    A good way to illustrate is to let the inner function, \(\cos 2x = u\), and find dy/du and du/dx. Their product is dy/du * du/dx = dy/dx, which is what we want to find. First, let's find dy/du. \( \displaystyle y = \sin u \) \( \displaystyle \frac{\text{d}y}{\text{d}u} = \frac{\text{d}}{\text{d}u} \sin u \) \( \displaystyle \frac{\text{d}y}{\text{d}u} = \cos u \) Now, du/dx. \( \displaystyle u = \cos 2x \) \( \displaystyle \frac{\text{d}u}{\text{d}x} = \frac{\text{d}}{\text{d}x} \cos 2x \) \( \displaystyle \frac{\text{d}u}{\text{d}x} = - 2 \sin 2x \) dy/dx = dy/du * du/dx. It's kind of like cancelling out, in essence. \( \displaystyle \frac{\text{d}y}{\text{d}u} \cdot \frac{\text{d}u}{\text{d}x} = \left(\cos u\right) \left( -2 \sin 2x \right) \) \( \displaystyle \frac{\text{d}y}{\text{d}x} = -2 \sin 2x \cos ( \cos 2x ) \)

    • one year ago
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