## cannotdoosmath 2 years ago Find the derivative: y= sin(cos(2x))

1. AccessDenied

Are you familiar with the Chain Rule for differentiation?

2. cannotdoosmath

Yes

3. cannotdoosmath

4. cannotdoosmath

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

5. AccessDenied

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

6. cannotdoosmath

derivative is cos

7. AccessDenied

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

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