Hi everyone,
I'm in Calc BC, but this year my teacher seems to make a lot of mistakes. I just wanted to ask a clarifying question about a second derivative of the function below:
$\displaystyle \\f(x) = \frac{x^8}{x^4}
\\f^{''}(x) = ?$
I assumed like many of my classmates that we could just do this:
$\displaystyle \\f(x) = \frac{x^8}{x^4} = x^{4}
\\f^{'}(x) = 4x^3
\\ f^{''}(x) = 12x^2$
But she said that we had to take into account the original function, and that since there was the $\displaystyle x^4$ as the denominator, we had to include the extra domain restriction of $\displaystyle x \neq 0$. This doesn't really make sense to me, because I figure you can just reduce it and that shouldn't be a problem. Obviously if the bottom was $\displaystyle x^41$ I would've included the restriction $\displaystyle x \neq 1$.
Am I right to think this?
Thanks,

Adam