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I have no idea how to do this problem I am suppose to find the derivative of $y=\cos(a^3 + x^3)$ but I do not really know how to start this problem. At first I thought I could use the sum rule and just make it $3a^2$ and $3x^2$ but for some reason I do not think that is correct, possibly because I am solving for $x$ and for some rule I can not get the derivative of a like that.

Anyways what I tried was using the product rule which gave me $\cos(a^3 + x^3)\prime -\sin(a^3 + x^3)$ not sure if that makes sense but I attempted to get the derivative of $(a^3 + x^3)$ using the chain rule and I got completely the wrong answer. I ended up with this abomination, $3\cos(a+x)^2 -\sin(a^3 + x^3)$

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    It is the Chain Rule. Let $u=a^3+x^3$. Then $y=\cos u$. Note that since $a$ is assumed to be a constant, $\frac{du}{dx}=3x^2$. I think the rest of the Chain Rule has been explained to you. You should end up with $-3x^2\sin(a^3+x^3)$, or something equivalent to that.2011-09-22
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    This is just a relatively straightforward application of your previous question. I'd suggest looking at [Jonas' answer](http://math.stackexchange.com/questions/66724/the-chain-rule-how-to-interpret-it/66744#66744) again.2011-09-22
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    Is $a^3+x^3$ the same as $(a+x)^3$? to me it is when I put in numbers for a and x.2011-09-22
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    @Jordan: No, it's not the same. And it's not the same when you put in numbers, either. If $a=x=1$, then $a^3+x^3 = 1^3+1^3 = 1+1 = 2$; but $(a+x)^3 = (1+1)^3 = 2^3 = 8$. Is $2$ the same as $8$?2011-09-22
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    No it isn't. Almost any choice of $a$ and $x$ will show that. Take for example $a=1$, $x=2$.2011-09-22
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    I still can't get the correct answer, I am getting $-3sin(a^3+x^3)(3a^2 + 3x^2)$2011-09-22
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    @Jordan: Where did the first $3$ come from? And you are still treating $a^3$ as if $a$ were a variable. It's a *constant*, so its derivative is $0$.2011-09-22
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    Well I did 2 for a and x but I guess that is wrong because they are both 2, but it was something I could do in my head.2011-09-22
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    As a second point, I'm not a fan of Newtonian notation to indicate a derivative, e.g. $f'(x)$, and I prefer Leibniz's, e.g. $df/dx$, because it is difficult to tell which variable to use when calculating a derivative like, $\cos(a^3 + x^3)'$. In first semester calculus, this is easier as it is likely to be $x$. But, Leibniz's notation leaves less room for ambiguity, and that can be a life saver when doing these. However, professors insist on using Newton's form, so you have to learn it, also. But, I'd change it to the other form when working these.2011-09-22
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    What do you mean "you did 2 for a and x"? Again: $a$ is a constant, $a^3$ is just a constant, the derivative of a constant is $0$, no matter how you write it. The derivative of $\pi^8$ is $0$, not $8\pi^7$, because $\pi^8$ is... a constant. The derivative of $a^3$ is $0$, because $a^3$ is a constant.2011-09-22
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    The $a$ is supposed to be a constant, so $a^3$ is a constant, the derivative of a constant with respect to $x$ is $0$. By the way, it is dangerous to put the $3a^2+3x^2$ (which should be $3x^2$) at the end like you did, since it may look as if you are taking the sine of $[(a^3+x^3)(3x^2)]$. Better is $(3x^2)(-\sin(a^3+x^3))$ or more simply $-3x^2\sin(a^3+x^3)$.2011-09-22
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    Yeah I forgot that a was a constant, I don't know how I overlooked that. I do that a lot when I am working with problems.2011-09-22

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Please don't write things like "$+-\sin(a^3+x^3)$"; they tend to confuse you later on. If you must put the addition and the negative sign, use parentheses, like $+(-\sin(a^3+x^3))$.

Next: $a$ is a constant; the variable is $x$. Constants have zero derivative. So you don't get $3a^2$ and $3x^2$.

Finally: the derivative of $y=\cos(a^3+x^3)$. This is a composition: if you write it as $f(g(x))$, then $g(x) = a^3+x^3$ (first you cube $x$ and add $a^3$ to it) and $f(u)=\cos u$ (then you take the cosine of whatever you have).

So $g'(x) = (a^3)' + (x^3)' = 0 + 3x^2 = 3x^2$ (again, $a$ is a constant).

On the other hand, $f'(u) = -\sin(u)$.

So, using the Chain Rule that says that $$\Bigl( f\bigl(g(x)\bigr)\Bigr)' = f'\bigl(g(x)\bigr)g'(x),$$ we have $$\begin{align*} \left(\cos(a^3+x^3)\right)' &= \left(f'(a^3+x^3)\right)g'(x)\\ &= \left(-\sin(a^3+x^3)\right)\left(3x^2\right)\\ &= -3x^2\sin(a^3+x^3). \end{align*}$$

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So is y the function of x: i.e. $y(x)$? or y is the function of both x and a: i.e. $y(x,a)$?

If y is only the function of x, then you just need to use the chain rule to de the derivative:

$dy = -\sin(a^3+x^3)\cdot 3x^2 dx$

If y is the function for both x and a, then you should use chain rule on both variables:

$dy = -\sin(a^3 + x^3)\cdot (3a^2 da + 3x^2 dx)$

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    Whether the function is of one variable or two depends on the context. It is a kind of convention that $x$ is a variable and $a$ is a constant, but your observation does point out to me that the question asks for the derivative, but does not specify what derivative is required. However, $x$ is stated to be variable and $a$ constant.2011-09-22