Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

jkristia

  • 2 years ago

Problem 5A-3g, is the answer correct?

  • This Question is Closed
  1. jkristia
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm not able to get the same nice clean answer as in the solution, and I think the problem is a mistake in the given solution where he get \[\frac{dy}{dx} \frac{x}{\sqrt{1 - x^2}} = (1-x^2)^{-3/2} ??\] it seems like he forgot the \(x\) in the numerator I get \[=\frac{-x^2+x+1}{(1-x^2)\sqrt{1-x^2}}\] What am I missing, where is my mistake ?

  2. OBMD
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    First change the equation to \[\frac{ dy }{ dx}x(x-x ^{2})^{-\frac{ 1 }{ 2 }}\] Use the product rule you get \[(1)(1-x ^{2})^{-\frac{ 1 }{ 2 }}+x(-\frac{ 1 }{ 2 })(1-x ^{2})^{-\frac{3 }{ 2 }}(-2x)\] this can be simplified to \[\frac{ 1-x ^{2} }{(1-x ^{2})^{\frac{3}{2}} }+\frac{ x ^{2} }{(1-x ^{2})^{\frac{3}{2}} }\] which equals \[\frac{ 1 }{ (1-x ^{2} )^{\frac{ 3 }{ 2 }}}\] or\[(1-x ^{2})^{\frac{ -3 }{ 2 }}\]

  3. OBMD
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Note that \[(1-x ^{2})^{-\frac{ 1 }{ 2 }}=\frac{ 1 }{ (1-x ^{2})^{\frac{ 1 }{ 2 }} }=\frac{ 1-x ^{2} }{ (1-x ^{2})^{ \frac{ 3 }{ 2 } }}\]

  4. jkristia
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh man.... you are right :). I have to go back and check my calculation, but I think my silly mistake was to multiply by (-2), not (-2x), arghhh... sometime you just go blind staring at a problem.

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy