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 $f(x)$.


Using the power rule again, we find the second derivative of $f(x)$.


Finally, since we know that $f′′(0.5)=3$, we set $x=0.5$ and solve for $a$.




  • 2
    $\begingroup$ 'a' is a constant. $\endgroup$ – Pedro Tamaroff Oct 31 '14 at 22:37
  • $\begingroup$ $\ a$ is a constant, it just multiplies the $\ x^3$ term $\endgroup$ – Mosk Oct 31 '14 at 22:38
  • $\begingroup$ $a$ is a constant (not dependent on $x$) and even if it were dependent on $x$, you would most definitely not get what you wrote. You would just get another term with derivative of $a$. The $a$ will never just disappear. $\endgroup$ – orion Oct 31 '14 at 22:41

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, $a=\frac{3}{11}$.

  • $\begingroup$ So a is a constant. Got it. How do you tell the difference between a constant and variable in a function? d/dx of a constant is 0, while d/dx of a variable, say x, is 1. Is it because of the order? Defined constant, undefined constant, variable? Or do you know this just because it is in standard form (for polynomials.) What about other kinds of functions? Thank you so much for the help. $\endgroup$ – Novice Polymath Nov 1 '14 at 3:19
  • $\begingroup$ @NovicePolymath You are evaluating $\dfrac{d}{dx}$, so $x$ is your 'changing' variable. It describes the rate of change with respect to $x$. Likewise $\dfrac{dy}{dt}$ describes the rate of change of $y$ with respect to $t$, so $t$ is your variable. $\endgroup$ – Landuros Sep 10 '18 at 12:03

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.


Your big mistake:


No, the value of $a$ does not depend on the value of $x$, so that


You could consider the derivative on $a$,


but this is irrelevant for the problem on hand.

  • 1
    $\begingroup$ Also note a second mistake when applying the product rule, $d(3ax^3)/dx=3\,da/dx\,x^3+3a\,dx^3/dx$. $\endgroup$ – Yves Daoust Sep 10 '18 at 12:01

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