Derivative using product rule and chain rule 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, $3\cos(a+x)^2  -\sin(a^3 + x^3)$
 A: 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
$$\begin{align*}
\left(\cos(a^3+x^3)\right)' &= \left(f'(a^3+x^3)\right)g'(x)\\
&= \left(-\sin(a^3+x^3)\right)\left(3x^2\right)\\
&= -3x^2\sin(a^3+x^3).
\end{align*}$$
A: 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:
$dy = -\sin(a^3 + x^3)\cdot (3a^2 da + 3x^2 dx)$
