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Homework Help: Calc - Derivatives and Differentiation of Logs

  1. Apr 21, 2010 #1
    1. The problem statement, all variables and given/known data
    I have a few problems that are giving me some trouble:

    1. Take the derivative of xe-4x

    2. Find dy/dx and evaluate the slope for the curve ey^3 - 2x4 + y2 = 3 at (8,0)

    3. Find dy/dx and evaluate the slope for the curve e-y - 4 = x2 + 1 at (-2,2)

    2. Relevant equations


    3. The attempt at a solution

    1. I'm OK with taking these types of derivatives, but the x out in front is throwing me off. I'm not quite sure what to do with it. For now, I have this as my answer: -4xe-4x

    2. I'm not going to type out every step, but I took d/dx of both sides and eventually came up with dy/dx(3yey^3 + 2y) = 8x3

    Then I solved for dy/dx to get 8x3 / (3yey^3 + 2y)

    The only thing that gives me reason to believe my answer is incorrect is the fact that when I plug in the values for x and y from the coordinate given, I get 4096 / 0, which would mean the slope is undefined. For some reason this doesn't seem right.

    3. I took d/dx of both sides and eventually got -e-ydy/dx = 2x

    Then I solved for dy/dx to get 2x / (-e-y)

    Then I plugged in the coordinate values and got a slope of 29.6. Is this correct? I wasn't sure if it was OK to move the terms as such in the beginning: e-y - x2 = 5
  2. jcsd
  3. Apr 21, 2010 #2
    1. You must use the product rule With f = x and g = e-4x.

    2. Are you using implicit differentiation? If not you should.

    3. You can move the terms, you are just simplifying the equation. Personally I would isolate the y variable, take the ln of both sides and then differentiate.
  4. Apr 21, 2010 #3


    Staff: Mentor

    Nope. You need to use the product rule.
    It doesn't look like you used the chain rule correctly. E.g., d/dx(ey2) = ey2 * d/dx(y2) = ey2 * 2y*y'.
    This one looks OK. Sure you can move terms around from one side to the other of an equation and then take the derivative.
    Last edited: Apr 22, 2010
  5. Apr 21, 2010 #4
    Really? Why wouldn't the x just go away? For example, d/dx (ex) = e ?
  6. Apr 21, 2010 #5


    Staff: Mentor

    d/dx(k*f(x)) = k *d/dx(f(x)), so d/dx(ex) = e*d/dx(x) = e

    But the problem I referred to had xe-4x, so the product rule has to be used. Although e is a constant, e-4x is a function of x.
  7. Apr 21, 2010 #6
    Oh, OK I see.

    I'm not sure if I used the product rule correctly in this case. I'm still a bit shaky with some of these derivatives.

    Here is what I did:

    f '(x) = 1 * e-4x + x * -4e-4x
    Then simplifying to get e-4x - 4xe-4x as my final answer.
  8. Apr 21, 2010 #7


    Staff: Mentor

    That's more like it.
  9. Apr 21, 2010 #8
    OK, great. I'm still having major trouble with the second problem. This ey3 is driving me insane.

    Did you say that I need to use the chain rule on it? I'm really confused about this.

    I'm don't really understand the example of the chain rule you posted earlier. To me, it looks like it suggests that d/dx (ey3) = ey3 * 2y * 2y ? Also, for this type of problem, how would you account for adding dy/dx variables in, as in implicit differentiation? Would you just proceed as normal?
    Last edited: Apr 21, 2010
  10. Apr 21, 2010 #9
    dy/dx[ey3] = ?

    dy/dx [ey] = ey * dy/dx Does this help you?
  11. Apr 21, 2010 #10
    d/dx (ey3) = ey3 * 2y * 2y is not correct.

    You know that d/du[eu] = eu*u'. So what is the derivative of y3? Namely what is d/dx [y3] =? Then you can just multiply that by ey3 and that is your answer. There will be a dy/dx in your answer.
    Last edited: Apr 21, 2010
  12. Apr 21, 2010 #11

    Is it (3y2 * dy/dx) * ey3, then? But isn't this not the chain rule?
  13. Apr 21, 2010 #12
    y is an implicit function of x, so when you differentiate d/dx e^(y^3) you have to use the chain rule for e ^ ( y^3 ) [the ouside e^__ ] and then the chain rule again for y^3
    Last edited: Apr 21, 2010
  14. Apr 21, 2010 #13

    Yes that is the correct answer. It is the chain rule if you let u = ey3 then by the chain rule du/dx = du/dy * dy/dx.
    du/dy = ey3*(3y2) and dy/dx = dy/dx.
  15. Apr 21, 2010 #14
    Ok, now I'm stuck again, but only on something minor.

    I now have 3y2(dy/dx) * ey3 + 2y(dy/dx) = 8x3

    If I want to factor out a dy/dx, how can I combine the 3y2 with the ey3 ? Do I just lump them together, i.e. 3y2ey3 ?

    If so, my original answer was correct for the most part and I am still left with a slope of 4096 / 0 = und.
    Last edited: Apr 21, 2010
  16. Apr 21, 2010 #15
    3y2(dy/dx) * ey3 + 2y(dy/dx) = 8x3

    Like you would any other variable:


    So yes, the the order of multiplication does not matter and dy/dx is multiplying that argument.

    3y2(dy/dx) * ey3 = 3y2*(dy/dx) * ey3

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