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I would like to know the general formula for expressing $\dfrac{\partial^k}{\partial x^k} \left(\dfrac{f(x)}{g(x)}\right)$ in terms of derivatives of $f(x)$ and $g(x)$. I am stuck when trying to express $h^{(i)}(x)=\dfrac{\partial^i (g(x))^{-1}}{\partial x^i}$ in terms of derivatives of $g(x)$.

Attempt: Let $h(x)=\frac{1}{g(x)}$ and $h^{(i)}(x)=\dfrac{\partial^i h(x)}{\partial x^i}$. Then $$ \frac{\partial^k}{\partial x^k} \left(\frac{f(x)}{g(x)}\right) =\frac{\partial^k}{\partial x^k} \left(h(x)f(x)\right) =\sum_{i=1}^k \binom{k}{i}h^{(i)}(x)f^{(k-i)}(x) $$ How would I express $h^{(i)}(x)=\dfrac{\partial^i (g(x))^{-1}}{\partial x^i}$ in terms of derivatives of $g(x)$?

user103828
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1 Answers1

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Use the Faà di Bruno's formula:

$$ {d^n \over dx^n}h(g(x))=\sum \frac{n!}{m_1!\,1!^{m_1}\,m_2!\,2!^{m_2}\,\cdots\,m_n!\,n!^{m_n}}\cdot h^{(m_1+\cdots+m_n)}(g(x))\cdot \prod_{j=1}^n\left(g^{(j)}(x)\right)^{m_j} $$ and set $h(y)=\frac1y$.

draks ...
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