1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simplify the expression

  1. Apr 13, 2013 #1
    1. Simplify the derivative



    2. [itex] y = \frac{\sqrt{1-x^{2}}}{x}[/itex]



    3. [itex] f'(x) = \frac{x*\frac{-2x}{2\sqrt{1-x^{2}}}-\sqrt{1-x^{2}}}{x^{2}}=\frac{-x^{2}-\sqrt{1-x^{2}}}{x^{2}\sqrt{1-x^{2}}}[/itex] I just don't know how to simplify this further.
     
  2. jcsd
  3. Apr 13, 2013 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You simplified ##\displaystyle \ \frac{\displaystyle x\frac{-2x}{2\sqrt{1-x^{2}}}-\sqrt{1-x^{2}}}{x^{2}} \ ## incorrectly.
     
  4. Apr 13, 2013 #3
    So it would be -x[itex]^{2}[/itex] over the square root. So how do I find the second derivative of [itex]\frac{\frac{-x^{2}}{\sqrt{1-x^{2}}}-\sqrt{1-x^{2}}}{x^{2}}[/itex] Do I use the product rule or quotient rule. Which is easier?
     
    Last edited: Apr 14, 2013
  5. Apr 14, 2013 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Multiply by ##\displaystyle \ \frac{\sqrt{1-x^2}}{\sqrt{1-x^2}} \ .##
     
  6. Apr 14, 2013 #5
    I already did that and came out with [itex] \frac{-1}{x^{2}\sqrt{1-x^{2}}}[/itex]
     
  7. Apr 14, 2013 #6
    How do I take the second derivative?
     
  8. Apr 14, 2013 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You can use the quotient rule. To differentiate the denominator use the product rule. Alternatively, rewrite it as a product of two terms with negative exponents and just use the product rule.
     
  9. Apr 14, 2013 #8
    Here is what I have [itex]\frac{\frac{-x}{\sqrt{1-x^{2}}}-\frac{x^{3}}{(1-x^{2})^{3/2}}} x^{2} + \frac{2}{x^{3}}[/itex] The squared on the top is supposed to go with the x on the bottom.
     
  10. Apr 14, 2013 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You mean [itex]\displaystyle \ \ \frac{\displaystyle \frac{-x}{\sqrt{1-x^{2}}}-\frac{x^{3}}{(1-x^{2})^{3/2}}} {x^{2}} + \frac{2}{x^{3}}\ \ ?[/itex]

    I don't see how that can possibly be correct.

    Please show some steps.
     
  11. Apr 16, 2013 #10
    Here is how I got that answer [itex]\frac{\frac{-2x^{3}}{2(1-x^{2})^{3/2}}}+\frac{2x}{2\sqrt{1-x^{2}}}-\frac{2x}{\sqrt{1-x^{2}}}{x^{2}}-2\frac{\frac{-x^{2}}{\sqrt{1-x^{2}}}}-\sqrt{1-x^{2}}{x^{3}} = \frac{\frac{-2x^{3}}{2(1-x^{2})^{3/2}}}-\frac{x}{\sqrt{1-x^{2}}}{x^{2}}-2\frac{\frac{-x^{2}}{\sqrt{1-x^{2}}}}-\sqrt{1-x^{2}}{x^{3}} = \frac{\frac{-x}{\sqrt{1-x^{2}}}}-\frac{x^{3}}{(1-x^{2})^{3/2}}{x^{2}}-2\frac{\frac{-x^{2}}{\sqrt{1-x^{2}}}}-\sqrt{1-x^{2}}{x^{3}}=\frac{\frac{-x}{\sqrt{1-x^{2}}}}-\frac{x^{3}}{(1-x^{2})^{3/2}}{x^{2}}+\frac{2}{x^{3}}[/itex] That is how I got the answer. The + sign in the first part should be in the numerator and the x^2 should be in the denominator. Same thing with the sign in the second half of the first part and the sqrt should be in the numerator with the x^3 on the bottom, same thing in the second part and third part and last part. Sorry it got messed up.
     
  12. Apr 16, 2013 #11

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It doesn't need to be that complicated. Write the first derivative as ##-x^{-2}(1-x^2)^{-\frac 12}## and differentiate using the product rule. Hint: every term in the answer ought to have a factor like ##(1-x^2)^{n-\frac 12}##, some integer n. I'd guess that's how SammyS knew your answer could not be right.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Simplify the expression
Loading...