# Thread: Domain of a second derivative

1. ## Domain of a second derivative

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:

$\\f(x) = \frac{x^8}{x^4}
\\f^{''}(x) = ?$

I assumed like many of my classmates that we could just do this:

$\\f(x) = \frac{x^8}{x^4} = x^{4}
\\f^{'}(x) = 4x^3
\\ f^{''}(x) = 12x^2$

But she said that we had to take into account the original function, and that since there was the $x^4$ as the denominator, we had to include the extra domain restriction of $x \neq 0$. 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 $x^4-1$ I would've included the restriction $x \neq 1$.

Am I right to think this?

Thanks,

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If you have, for example, $f(x) = \frac{x^2}{x}$, x cannot equal 0. You can't cancel out the denominator when x is 0, the function is not continuous there.
4. Why must we think of $\frac{x^2}{x}$ and $x$ as two different things? We were always taught that you can simplify the former into the latter. Is there any specific reason, or is it something I should just accept?
Why must we think of $\frac{x^2}{x}$ and $x$ as two different things? We were always taught that you can simplify the former into the latter. Is there any specific reason, or is it something I should just accept?
NO, you weren't taught that. You were taught that $\frac{x^2}{x}= x$ for all x except x= 0. If you thought differently, you were not paying attention. You cannot divide by 0. The function $f(x)= x$ has natural domain "all real x". The function $f(x)= \frac{x^2}{x}$ has domain "all real x except x= 0".