$\frac{\mathrm d}{\mathrm{d}x}3ax^3$ not equal to $9x^2$? Okay, so I'm doing some Khan Academy stuff and they ask me to take $\frac{\mathrm d}{\mathrm{d}x}$ of this function: $f(x)=3ax^3+ax^2$ where $f''(0.5)=3$. They say that the derivative of $f(x)$ is $9ax^2 + 2ax$. This doesn't make sense to me though.
The constant rule states that $\frac{\mathrm d}{\mathrm{d}x}(cf)=cf'$.
The product rule states that $\frac{\mathrm d}{\mathrm{d}x}(fg)=fg'+f'g$.
$3$ is a constant.  $a$ is not.  $\frac{\mathrm d}{\mathrm{d}x}(x^3)=3x^2$.
So $\frac{\mathrm d}{\mathrm{d}x}(3ax^3)=3\cdot\frac{\mathrm d}{\mathrm{d}x}(a)\cdot\frac{\mathrm d}{\mathrm{d}x}(x^3)=3\cdot 1\cdot\frac{\mathrm d}{\mathrm{d}x}(x^3)$, where $3$ is a constant and $x^3$ is not.
$3\frac{\mathrm d}{\mathrm{d}x}(x^3)=3\cdot3x^2=9x^2$. 
Khan Academy says that $\frac{\mathrm d}{\mathrm{d}x}(3ax^3+ax^2)=9ax^2+2ax$.  Why all the $a$'s?
So where am I going wrong?  Is $3$ a constant of $a$, and so when you take the derivative of $3a$ (which is $3$) and multiply it by $x^3$ you need to use the product rule? Or do you do $\frac{\mathrm d}{\mathrm{d}x}(3)\cdot\frac{\mathrm d}{\mathrm{d}x}(ax^3)$? $\frac{\mathrm d}{\mathrm{d}x}(ax^3)$ with the product rule? 
Moreover, they say that $f''(0.5)=18a(0.5)+2a$.
Where did the $x$ go?  If $x=0.5$ then $2ax$ should be equal to $a\cdot 2\cdot 0.5=a$.
I'll copy the entire problem down for you below.

If $f(x)=3ax^3+ax^2$ and $f′′(0.5)=3$, then what is the value of $a$?
a. $3/11$, b. $1/4$, c. $3/7$, d. $1/2$, e. None of the above
We need to find the second derivative of the function $3ax^3+ax^2$.  The power rule is given by $\frac{\mathrm d}{\mathrm{d}x}(x^n)=nx^{n−1}$.  Using the power rule, we find the first derivative of $f(x)$.
$f'(x)=9ax^2+2ax$
Using the power rule again, we find the second derivative of $f(x)$.
$f''(x)=18ax+2a$
Finally, since we know that $f′′(0.5)=3$, we set $x=0.5$ and solve for $a$.
$f''(0.5)=18a(0.5)+2a$
$3=11a$
$a=3/11$
 A: In this case, $x$ is your $\textbf{variable}$ while $a$ is some $\textbf{constant}$, or fixed point. 
Therefore, the first derivative of the function $f(x) = 3ax^3+ax^2$ is $$f'(x) = 9ax^2+2ax$$ because $3a$ is a constant in $f(x)$ and $1*a$ is a constant in $f(x)$ as well. Next, we take the derivative of $f'(x)$, which gives $$f''(x) = 18ax+2a.$$ Again, I want to emphasis the point that $a$ is a constant number. Since we know that $f''(.5)=3$, we plug in $x=.5$ and set $f''(.05)$ equal to $3$ and solve. Therefore, we have that $$3=18a(.5)+2a = 9a+2a=11a$$ which gives $$3=11a.$$ Therefore, $a=\frac{3}{11}$. 
A: You want to find  $$ \frac {d}{dx} (3ax^3+ax^2)$$
That means you are taking derivative with respect to $x$,  therefor other letters are considered constants.
Usually they denote constants with the beginning letters of alphabet 
 such that $a,b,c$ and variables with the $x,y,z$ so that is another hint that $a$ is a constant in this problem.  
A: Your big mistake:
$$\frac{da}{dx}=\color{red}1.$$
No, the value of $a$ does not depend on the value of $x$, so that
$$\frac{da}{dx}=\color{green}0.$$

You could consider the derivative on $a$,
$$\frac{da}{da}=1$$
but this is irrelevant for the problem on hand.
