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I'm baffled by Pset 1G5(b). I've checked the answer and can't make sense of it.
1. I don't see how all terms but the one provided contain a 0factor.
2. I don't understand the reasoning behind the following:
"If u =x^p, then u^p = p!"
Does the p here refer only to the pth derivative of u, or also to the pth power of x?
FWIW, I calculated the third derivative of 1G5(a), and found the following. (The 3rd derivative is not given in the solution sheet.):
y''' = u'''v + 3u''v' + 3u'v'' +uv'''
Thank you!
 one year ago
 one year ago
I'm baffled by Pset 1G5(b). I've checked the answer and can't make sense of it. 1. I don't see how all terms but the one provided contain a 0factor. 2. I don't understand the reasoning behind the following: "If u =x^p, then u^p = p!" Does the p here refer only to the pth derivative of u, or also to the pth power of x? FWIW, I calculated the third derivative of 1G5(a), and found the following. (The 3rd derivative is not given in the solution sheet.): y''' = u'''v + 3u''v' + 3u'v'' +uv''' Thank you!
 one year ago
 one year ago

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van1234Best ResponseYou've already chosen the best response.0
Yeah the notation is a little bit funky here. \[u = x ^{p}\] is just a function u where p is just an exponent. The other one which is: \[u ^{(p)}\] (notice the parentheses) Basically means that you are taking the derivative of the function u, p times. The higher derivatives are surrounded by ( ) so you can distinguish them from just normal exponents. And from the looks of it, you third derivative looks good.
 one year ago
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