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!

Derivative of exponential function

  1. Dec 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Use implicit differentiation to find dy/dx.

    2. Relevant equations

    xey - 10x + 3y = 0

    3. The attempt at a solution

    = [xey + ey(y)'] - (10x)' + (3y)' = 0

    = xey + ey(y') - 10 + 3(y') = 0

    = y' (ey + 3) = 10 - xey

    = y' = 10 - xey/ ey + 3y

    However, my book says the answer: 10 - ey/ xey + 3y.

    What am I doing wrong?

    Thanks!
     
  2. jcsd
  3. Dec 7, 2012 #2

    Mark44

    Staff: Mentor

    The first thing wrong is starting the line below with =. By putting that = at the start of each line, you are more likely to lose track of the two sides of the equation you started with.
    The more serious mistake is above. You didn't take the derivative of xey correctly. d/dx(xey) = ey + xey*y'. Do you see why?
    No, I believe their answer is (10 - ey)/(xey + 3). Note the parentheses, and the 3 instead of the 3y that you had.
     
  4. Dec 7, 2012 #3
    About taking the derivative wrongly, my question is: How does that differ from y= xe^x - e^x?

    Because I was trying to follow the same thinking when solving the problem in question.

    This is how I solved it ( and I got it right):

    = xe^x + e^x (x)' - e^x(x)'
    = xe^x + e^x - e^x
    = e^x ( x -1 + 1)
    = xe^x

    Thanks!
     
    Last edited: Dec 7, 2012
  5. Dec 7, 2012 #4

    Mark44

    Staff: Mentor

    Don't start your work with =.
    In this problem,
    d/dx(xex) = x *d/dx(ex) + d/dx(x) * ex = xex + 1*ex = ex(x + 1)

    The previous problem is different because you need to use the chain rule.
    d/dx(xey) = x * d/dx(ey) + d/dx(x) * ey
    = x * d/dy(ey) * dy/dx + d/dx(x) * ey = xey*y' + 1*ey = ey(xy' + 1)
     
  6. Dec 7, 2012 #5
    Ok, I am going to try to solve some other problems to practice it.
     
    Last edited: Dec 7, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook