Vtyjkyuk

Okay, so I'm doing some Khan Academy stuff and they ask me to take $fracmathrm dmathrmdx$ 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 $fracmathrm dmathrmdx(cf)=cf'$.
The product rule states that $fracmathrm dmathrmdx(fg)=fg'+f'g$.

$3$ is a constant. $a$ is not. $fracmathrm dmathrmdx(x^3)=3x^2$.

So $fracmathrm dmathrmdx(3ax^3)=3cdotfracmathrm dmathrmdx(a)cdotfracmathrm dmathrmdx(x^3)=3cdot 1cdotfracmathrm dmathrmdx(x^3)$, where $3$ is a constant and $x^3$ is not.

$3fracmathrm dmathrmdx(x^3)=3cdot3x^2=9x^2$.

Khan Academy says that $fracmathrm dmathrmdx(3ax^3+ax^2)=9ax^2+2ax$. 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 $fracmathrm dmathrmdx(3)cdotfracmathrm dmathrmdx(ax^3)$? $fracmathrm dmathrmdx(ax^3)$ 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 $acdot 2cdot 0.5=a$.

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 $fracmathrm dmathrmdx(x^n)=nx^n−1$. 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 $a$.

$f''(0.5)=18a(0.5)+2a$

$3=11a$

$a=3/11$

In this case, $x$ is your $textbfvariable$ while $a$ is some $textbfconstant$, 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, $a=frac311$.

You want to find $$ frac ddx (3ax^3+ax^2)$$

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:

$$fracdadx=colorred1.$$

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

$$fracdadx=colorgreen0.$$

You could consider the derivative on $a$,

$$fracdada=1$$

but this is irrelevant for the problem on hand.