<s>[INST] How to Calculate the derivative of $\frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}}$ I'm sorry to bother you with this easy problem. But I'm working alone and totally confused.
It is the 1378th Problem from "Problems in Mathematical Analysis" written by Demidovich. The standard answer is
$$\frac{60(1+x)^{99}(1+6x)}{(1-2x)^{41}(1+2x)^{61}}$$
But when I followed the rutine
$$\frac{d(P(x)/Q(x))}{dx}=\frac{\frac{dP(x)}{dx}Q(x)-P(x)\frac{dQ(x)}{dx}}{Q(x)^2}$$
I got
$$\frac{100(1+x)^{99}(1-2x)^{40}(1+2x)^{60}-(1+x)^{100}(-80(1-2x)^{39}+120(1+2x)^{59})}{(1-2x)^{80}(1+2x)^{120}}$$
and finally
$$\frac{60(1+x)^{99}P(x)}{(1-2x)^{41}(1+2x)^{61}}$$
$$P(x)=1-4x^2-\frac{3(1+x)}{(1-2x)^{39}}-\frac{2(1+x)}{(1+2x)^{59}}$$
Would you tell me what mistake I made. Best regards.
 [/INST] You can just take the derivative of the logarithm:
$$y = \frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}} \implies \ln y = 100\ln(1+x)-40\ln(1-2x)-60\ln(1+2x)$$
$$\frac{y'}{y} = (\ln y)' = \frac{100}{1+x} + \frac{80}{1-2x} - \frac{120}{1+2x}$$
\begin{align}y' &= y\left(\frac{100}{1+x} + \frac{80}{1-2x} - \frac{120}{1+2x} \right)\\
&= \frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}}\left(\frac{100}{1+x} + \frac{80}{1-2x} - \frac{120}{1+2x} \right)\\
&= \frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}}\frac{100(1-4x^2)+80(1+x)(1+2x)-120(1+x)(1-2x)}{(1+x)(1-2x)(1+2x)}\\
&= \frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}}\frac{60(1+6x)}{(1+x)(1-2x)(1+2x)}\\
&= \frac{60(1+x)^{99}(1+6x)}{(1-2x)^{41}(1+2x)^{61}}
\end{align}
</s>