## cannotdoosmath Group Title Find the derivative: y= sin(cos(2x)) one year ago one year ago

1. AccessDenied Group Title

Are you familiar with the Chain Rule for differentiation?

2. cannotdoosmath Group Title

Yes

3. cannotdoosmath Group Title

4. cannotdoosmath Group Title

so sin(2x) * d/dx (sin(2x)?

5. AccessDenied Group Title

What is the general set up for the Chain Rule? y = f(g(x)) What is the derivative of this general function?

6. cannotdoosmath Group Title

derivative is cos

7. AccessDenied Group Title

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.

8. AccessDenied Group Title

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. $$\displaystyle \frac{\text{d}y}{\text{d}u} \cdot \frac{\text{d}u}{\text{d}x} = \left(\cos u\right) \left( -2 \sin 2x \right)$$ $$\displaystyle \frac{\text{d}y}{\text{d}x} = -2 \sin 2x \cos ( \cos 2x )$$