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AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.1Are you familiar with the Chain Rule for differentiation?

cannotdoosmath
 one year ago
Best ResponseYou've already chosen the best response.0Help me along though please

cannotdoosmath
 one year ago
Best ResponseYou've already chosen the best response.0so sin(2x) * d/dx (sin(2x)?

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.1What is the general set up for the Chain Rule? y = f(g(x)) What is the derivative of this general function?

cannotdoosmath
 one year ago
Best ResponseYou've already chosen the best response.0derivative is cos

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.1Hmm... 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.

AccessDenied
 one year ago
Best ResponseYou've already chosen the best response.1A 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. \( \displaystyle \frac{\text{d}y}{\text{d}u} \cdot \frac{\text{d}u}{\text{d}x} = \left(\cos u\right) \left( 2 \sin 2x \right) \) \( \displaystyle \frac{\text{d}y}{\text{d}x} = 2 \sin 2x \cos ( \cos 2x ) \)
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