anonymous
  • anonymous
Find the derivative: y= sin(cos(2x))
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AccessDenied
  • AccessDenied
Are you familiar with the Chain Rule for differentiation?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Help me along though please

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so sin(2x) * d/dx (sin(2x)?
AccessDenied
  • AccessDenied
What is the general set up for the Chain Rule? y = f(g(x)) What is the derivative of this general function?
anonymous
  • anonymous
derivative is cos
AccessDenied
  • 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.
AccessDenied
  • 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. \( \displaystyle \frac{\text{d}y}{\text{d}u} \cdot \frac{\text{d}u}{\text{d}x} = \left(\cos u\right) \left( -2 \sin 2x \right) \) \( \displaystyle \frac{\text{d}y}{\text{d}x} = -2 \sin 2x \cos ( \cos 2x ) \)

Looking for something else?

Not the answer you are looking for? Search for more explanations.