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Derivative of y = cos(a^3 + x^3)

  1. Jul 24, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the derivative of the function y = cos(a^3 + x^3).


    2. Relevant equations
    - Chain rule
    - Power rule


    3. The attempt at a solution
    This is driving me insane.. so here's what I have.

    y = cos(a^3 + x^3)

    y' = -sin(a^3 + x^3) * d/dx (a^3 + x^3)

    y' = -sin(a^3 + x^3) * (3a^2 + 3x^2

    y' = -(3a^2)(sin(a^3 + x^3)) - (3x^2)(sin(a^3 + x^3))

    I would like to think that that's where this ends.. but my book claims the result is

    y' = -(3x^2)(sin(a^3 + x^3))

    How? Where the heck does that entire -(3a^2)(sin(a^3 + x^3)) term go!?
     
    Last edited by a moderator: Jul 24, 2008
  2. jcsd
  3. Jul 24, 2008 #2
    Hence, dy/dx not dy/dx + dy/da!
     
  4. Jul 24, 2008 #3
    Ah! Ok that makes sense on an intuitive level.. can you expand that a bit though? Does a simply turn into 0 and thus kill that whole term containing a as a factor?

    Thanks!

    Edit:
    By the way.. I stay far, far away from the dy/dx notation wherever possible. I just don't quite understand it.. for instance, I cringe when I see this in my book:

    [tex]\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]

    if

    [tex]\frac{dx}{dt}\neq0[/tex]

    for example.. I much prefer something like

    F'(x) = f'(g(x)) * g'(x).

    Granted, the two examples are not equivalent for the purposes of comparison.. but it gets the point across. I suppose that explains why Leibnitz' notation throws me off :/
     
    Last edited: Jul 24, 2008
  5. Jul 24, 2008 #4
    [tex]y'=\frac{dy}{dx}[/tex]

    Means to take the derivative of y with respects to x. Basically, only take the derivative of terms that include x.

    [tex]y=ax+a[/tex]

    You could easily mistake this as product rule, but it's not! (but you could if you wanted to) Anything that does not have an x variable associated with it, is a constant.

    [tex]\frac{dy}{dx}=a[/tex]

    or product rule (pointless)

    [tex]\frac{dy}{dx}=a\frac{dy}{dx}x+x\frac{dy}{dx}a+\frac{dy}{dx}a=a\cdot1+x\cdot0+0=a[/tex]

    [tex]y'=\frac{dy}{dt}[/tex]

    Simply means to take the derivative of y with respects to t or time. You can use any variable you want.
     
  6. Jul 24, 2008 #5

    HallsofIvy

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    No. d/dx(a^3+ x^3)= 3x^2. a^3 is a constant

     
  7. Jul 26, 2008 #6
    Sweet, that definitely helps :)

    While we're here.. is it against the rules or anything to ask a second question about a new problem in the same thread? Seems like a bit of a waste to start a new thread just for this next one..

    Here goes. If this is bad and frowned upon, let me know! (I understand, obviously, that as a general rule you don't want to do this.. but this one's so short I figure "what the hell"):

    [tex]y=xe^{-x^2}[/tex]

    I know the solution already, of course, but I'd appreciate a step by step "walkthrough".

    I got as far as this:

    [tex]
    y=xe^{-x^2}
    [/tex]

    [tex]
    y'=x\frac{d}{dx}e^{-x^2}+e^{-x^2}\frac{d}{dx}x
    [/tex]

    [tex]
    =xe^{-x^2}+e^{-x^2}
    [/tex]

    Also, can someone tell me why I can't seem to make a new line when writing tex? "\\" is supposed to do it.. but does not :/

    Oh btw.. the reason for the last step above was the assumption that the following is true:

    [tex]
    \frac{d}{dx}e^x=e^x[/tex]
     
  8. Jul 26, 2008 #7
    [tex]\frac{d}{dx}e^x=e^x[/tex] is true because the derivative of x is 1 (by power rule = derivative of x is 1 * x^(1-1)). If, in place of x, there was a function based on x (for example, 2x + 1), you would have e^(2x+1) multiplied by the derivative of the function, resulting in 2e^(2x+1) by the Chain Rule.

    Having said that, there is a flaw in your walkthrough. The derivative of e^(-x^2) would be, by the combination of the Chain Rule and the Power Rule, would be -2xe^(-x^2). Hence, your final answer should come out to [tex]-2x^{2}e^{-x^2}+e^{-x^2}[/tex].
     
  9. Jul 26, 2008 #8

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    It's not "against the rules", but many people will not look at a thread when they know the initial question has been answered- so you may be ruducing the number of people who see your new question.

    Yes, that last line is true. But it is NOT true that
    [tex]{d e^{x^2}}{dx}= e^{-x^2}[/tex]
    You forgot the chain rule: multiply by the derivative of -x2.
     
  10. Jul 26, 2008 #9
    Hello there:

    Here is my solution, which matches your textbook's.

    [tex] y = \cos(a^3 + x^3) [/tex]

    Let: [tex] u = a^3 + x^3 [/tex]

    The Chain Rule: [tex] h'(x) = f'(g(x)) \times g'(x) [/tex]

    Breaking the original equation down:

    [tex] f(x) = \cos u [/tex], [tex] f'(x) = -\sin u[/tex]

    [tex] g(x) = u = a^3 + x^3 [/tex], [tex] g'(x) = 3x^2 [/tex] (From the Product Rule)

    Note: [tex] 3a^2 [/tex] is actually a constant due to the term [tex] a [/tex]. Recall that the derivative of any constant is 0.)

    Compiling each function:

    [tex] h'(x) = f'(g(x)) \times g'(x) [/tex]

    [tex] h'(x) = f'(u) x 3x^2 [/tex]

    [tex] h'(x) = -\sin(a^3 + x^3)(3x^2) [/tex]

    [tex] h'(x) = -3x^2 \sin(a^3 + x^3)[/tex]

    Hope this helps.
     
  11. Jul 26, 2008 #10
    Thanks everyone. Your posts helped a lot! What threw me off, I believe, is that e's exponent had itself an exponent.. but what I need to remember specifically is the bold part quoted above!

    Thanks again :)
     
  12. Jul 27, 2008 #11
    I know this is probably really late, but I'd like to point out to rocomath his/her mistake in using the Leibniz notation: you should have used [tex]\frac{d}{dx}[/tex], not [tex]\frac{dy}{dx}[/tex]. The first one means "take the derivative of" and the second means "the derivative of y with respect to x is".

    Just being nitpicky. =]
     
  13. Jul 27, 2008 #12
    And that actually helped me understand Leibniz a bit better! Thanks :)
     
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