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, $3\cos(a+x)^2 -\sin(a^3 + x^3)$

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 $$\begin{align*} \left(\cos(a^3+x^3)\right)' &= \left(f'(a^3+x^3)\right)g'(x)\\\ &= \left(-\sin(a^3+x^3)\right)\left(3x^2\right)\\\ &= -3x^2\sin(a^3+x^3). \end{align*}$$