• anonymous
I'm baffled by P-set 1G-5(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 0-factor. 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 1G-5(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!
OCW Scholar - Single Variable Calculus
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
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
I got my questions answered at in under 10 minutes. Go to now for free help!
  • anonymous
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.

Looking for something else?

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