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Homework Help: 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?


    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:




    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

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


    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.


    or product rule (pointless)



    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


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    Science Advisor

    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"):


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

    I got as far as this:




    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:

  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


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    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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