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

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

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

AccessDeniedBest ResponseYou've already chosen the best response.1
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

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

AccessDeniedBest ResponseYou've already chosen the best response.1
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

AccessDeniedBest ResponseYou've already chosen the best response.1
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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