Can we compute \begin{align} \frac{d^k}{dx^k} \frac{f(x)}{g(x)} \Big|_{x=0} \end{align}
where $f(x)$ and $g(x)$ satisfy
- $ g(x)\neq 0$
- $\frac{d^k}{dx^k} f(x) \Big|_{x=0}=0$ and $\frac{d^k}{dx^k} g(x) \Big|_{x=0}=0$ for $k$ odd.
Higher order derivative of a ratio
1
$\begingroup$
calculus
real-analysis
derivatives
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0The quotient rule could work, but a general formula would be rather cumbersome. Instead, you could write $\frac{f(x)}{g(x)}=f(x)(g(x))^{-1}$ and use the [general Leibniz rule](https://en.wikipedia.org/wiki/General_Leibniz_rule) to write a formula for the $k$th order derivative. – 2017-01-11
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0@user170231 I see. Do we have the property that if $ \frac{d^n}{dx^n} g(x)=0$ then $ \frac{d^n}{dx^n} \frac{1}{g(x)}=0$ – 2017-01-11
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0@Boby Certainly not. Consider $g(x) =x$ for a counterexample when $n \ge 2$. – 2017-01-11
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0@Glitch Since, this is so. I don't think Leibnitz rule is much use if want to use my condition.s – 2017-01-11