anonymous
  • anonymous
sin(x/2)^2 - cos(x/2)^2 a. compute df/dx b. rewite answer from a in terms of sinx and cosx c. using b rewirte the original function in term s of sinx and cosx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Since no one has answered yet, here's my best run at it. I am not sure what is needed for B. since by doing A. you will get B. The steps that I would do for A. are: Break it into two parts, do the df/dx of the sin^2 and then the df/dx of the cos^2. Then, each of these will need two chain rules each. The first being the du/dx where u = sin(x/2) and du^2/du = 2u. The second being du/dx where u = cos(x/2) and du^2/du = 2u. After you do the above steps, you will have to do another chain rule on the sin (the left side) and another on the cos (the right side). It's going to get messy but the solution will come out to: sin(x/2)cos(x/2) + sin(x/2)cos(x/2) = sin(x).
anonymous
  • anonymous
Hmm, after reading my reply even I am confused as to what I was trying to say :) Let me try this a little slower. Split the function into two terms and differentiate each term. I will work the left term and leave the right term for you to do as it is exactly the same steps. 1. Let u = sin(x/2) then we can substitute to make the chain rule easier to see. d/dx sin(x/2)^2 = (d(u^2)/du)*(du/dx), where u = sin(x/2) and then d(u^2)/du = 2u. Substitute back in gives us (d(u^2)/du)*(du/dx) = 2u(du/dx) = 2sin(x/2) * ( d/dx (sin(x/2) ). 2. 2sin(x/2) * ( d/dx (sin(x/2) ), we need to do another chain rule on the d/dx sin(x/2). So we let u = (x/2) this time. So we will have d(sin(u))/du * du/dx, d(sin(u))/du is the derivative of sin which is cos. Substitute back in and we now have 2sin(x/2) * cos(x/2) * d/dx(x/2). 3. The next step is to solve for d/dx(x/2). Pull out the 1/2 so we get d/dx(x) which is 1. 4. This gives us 2sin(x/2) * cos(x/2) * 1 for the left term. The right term will be the same thing. I hope this makes better sense then my last try. :)

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