1. Not finding help here? Sign up for a free 30min 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!

Implicit Differentiation problem

  1. Jul 9, 2007 #1
    1. The problem statement, all variables and given/known data

    Find dy/dx by implicit differentiation when it is known that y^2 + xsiny = 4

    2. Relevant equations



    3. The attempt at a solution

    2y dy/dt + xcosy dy/dt + siny = 0

    2y dy/dt + xcosy dy/dt = -siny

    dy/dt + dy/dt = -siny/2y/xcosy

    I'm sure I'm doing it wrong so I stopped right there. What am I doing it wrong and how do I solve it?

    Thanks
     
  2. jcsd
  3. Jul 9, 2007 #2
    you're looking for dy/dx not dy/dt o.o, but assuming by dt you mean dx then in the 2nd line of your attempt you should pull out the dy/dt and the answer will be easy from there.
     
  4. Jul 9, 2007 #3
    Is this the answer?

    dy/dx = -siny/2y+xcosy
     
  5. Jul 9, 2007 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It would be if you would parenthesize the denominator.
     
  6. Jul 9, 2007 #5
    dy/dx = -siny/(2y+xcosy) ? What's that's do
     
  7. Jul 9, 2007 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It distinguishes between -(siny/(2y))+(xcosy) and -siny/(2y+xcosy). Which are two quite different expressions. But look the same if you omit the parentheses.
     
    Last edited: Jul 9, 2007
  8. Jul 9, 2007 #7
    [tex]\frac{d}{dx}(y^2+x\sin(y)) = 0[/tex]

    [tex]=2y\left(\frac{dy}{dx}\right)+\left[\frac{d}{dx}x\cdot\sin(y)+\frac{d}{dx}\sin(y)\cdot x\right]=0[/tex]

    [tex]= 2y\left(\frac{dy}{dx}\right) + \left[\sin(y) + x\cdot\cos(y)\left(\frac{dy}{dx}\right)\right]= 2y\left(\frac{dy}{dx}\right) + \sin(y) + x\cos(y)\left(\frac{dy}{dx}\right) = 2y\left(\frac{dy}{dx}\right) + x\cos(y)\left(\frac{dy}{dx}\right) + \sin(y) = 0[/tex]

    [tex]= \left(\frac{dy}{dx}\right)(2y+x\cos(y)) + \sin(y) = 0[/tex]

    [tex]= \left(\frac{dy}{dx}\right)(2y+x\cos(y)) = -\sin(y)[/tex]

    [tex]\left(\frac{dy}{dx}\right) = \frac{-\sin(y)}{(2y+x\cos(y))}[/tex]

    Edit: I'm late by about 25 minutes. :biggrin:
     
  9. Jul 9, 2007 #8
    cool thanks
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Implicit Differentiation problem
Loading...