Taking the constant out in derivatives. How does it work and how do I notice that I am able to do it? I am currently trying to solve the derivative to this function:
$$
f(x)= \frac{x^3+9x+8}{2x^2}
$$
by using the quotient rule only and after simplifying to the best of my ability I get:
EDIT:

$$\frac{\frac{d}{dx}(x^3+9x+8)\cdot2x^2-(x^3+9x+8)\cdot\frac{d}{dx}(2x^2)}{(2x^2)^2} \Leftrightarrow $$
$$
\frac{(3x^2+9)\cdot2x^2-(x^3+9x+8)\cdot2(2x)}{2x^4}
$$
$$
\frac{6x^4+18x^2-4x^4+36^2+32x}{2x^4}
$$
$$
\frac{2x^4+54x^2+32x}{2x^4}
$$
I just realized that there is a plus sign after $2x^4$ in the numerator, so that I can't just cancel them. Although, I have figured it out now. :)

$$
54x^2+32x
$$
However, when I input the function into Symbolabs calculator I get the following from finding the derivative:
$$
\frac{x^3-9x-16}{2x^3}
$$
And when taking a look at the steps (since I made a mistake), I could see that they took the constant out before even applying the quotient rule.
How do I acknowledge this? I don't understand how it's possible to "randomly" take out one of the constants and why it would be from the denominator as well. I understand every step but that.
Thanks in advance! I really appreciate it :)

Any time an expression, abc, can be rewritten as (a)(bc), where a doesn't contain any variables, "a" is a constant. Here,
$$
\frac{x^3+9x+8}{2x^2}
$$
Can be rewritten and retains the same mathematical value as:
$$
\frac 1 2 \frac{x^3+9x+8}{x^2}
$$
Therefore, $\frac 1 2$ is a constant that can be pulled out.
