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I have seen people use the approximation $\frac{(1+x)^n}{(1+y)^m}\approx1+nx-my$ where x and y are close to 0. I know $(1+x)^n \approx 1+nx$, but not sure about the ratio to difference approximation. Can someone please give an derivation?

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    $\begingroup$ In the same way, that $(1+x)^n \approx 1 + nx$, $(1+y)^{-m} \approx 1 - my$. Now multiply and keep terms to first order only. $\endgroup$
    – NickD
    Jan 11 '20 at 21:49
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For a formal proof, it results from Taylor's formula at order $1$ for two variables:

Set $f(x,y)=\dfrac{(1+x)^n}{(1+y)^m}.\;$ Taylor's formula at order $1$ asserts that $$f(x,y)=f(0,0)+\frac{\partial f}{\partial x}(0,0)\,x+\frac{\partial f}{\partial x}(0,0)\,y+o\bigl(\|(x,y)\|\bigr).$$ Here, we have $$\begin{cases}\frac{\partial f}{\partial x}=\frac{n(1+x)^{n-1}}{(1+y)^m},&\text{whence } \;\frac{\partial f}{\partial x}(0,0)=n, \\[1.5ex] \frac{\partial f}{\partial y}=-\frac{m(1+x)^n}{(1+y)^{m-1}},&\text{whence } \;\frac{\partial f}{\partial y}(0,0)= -m, \end{cases}$$ so that Taylor's formula becomes $$f(x,y)=1+nx-my+o\bigl(\|(x,y)\|\bigr).$$

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