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.
 A: Hint:) If you are familiar with logarithm so write
$$y=\frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}}$$
and
$$\ln y=100\ln(1+x)-40\ln(1-2x)-60\ln(1+2x)$$
take derivative and simplify!
A: You can write those equations as
$$\Biggr(\frac{1+x}{1-2x}\Biggr)^{40}\Biggr(\frac{1+x}{1+2x}\Biggr)^{60}$$
And use multiplication rule
It is easier this way

$$\color{green}{\frac{d}{dx} }\Biggr(\color{red}{\Biggr(\frac{1+x}{1-2x}\Biggr)^{40}}\color{blue}{\Biggr(\frac{1+x}{1+2x}\Biggr)^{60}}\Biggr)$$$$$$$$\color{red}{\Biggr(\frac{1+x}{1-2x}\Biggr)^{40}}\color{green}{\frac{d}{dx}}\color{blue}{\Biggr(\frac{1+x}{1+2x}\Biggr)^{60}}+\color{green}{\frac{d}{dx}}\color{red}{\Biggr(\frac{1+x}{1-2x}\Biggr)^{40}}\color{blue}{\Biggr(\frac{1+x}{1+2x}\Biggr)^{60}}$$$$$$$$\color{blue}{60 \Biggr(\frac{1+x}{1+2x}\Biggr)^{59}\Biggr(\frac{-1}{(1+2x)^2}\Biggr)}\color{red}{\Biggr(\frac{1+x}{1-2x}\Biggr)^{40}}+\color{red}{40\Biggr(\frac{1+x}{1-2x}\Biggr)^{39}\Biggr(\frac{3}{(1-2x)^2}\Biggr)}\color{blue}{\Biggr(\frac{1+x}{1+2x}\Biggr)^{60}}$$$$$$$$-60\frac{(1+x)^{99}}{(1+2x)^{\color{blue}{61}}(1-2x)^{\color{red}{40}}}+120\frac{(1+x)^{99}}{(1-2x)^{\color{red}{41}}(1+2x)^{\color{blue}{60}}}$$$$$$$$\frac{60(1+x)^{99}(\color{red}{-1+2x+2+4x})}{(1+2x)^{61}(1-2x)^{41}}$$$$$$$$\frac{60(1+x)^{99}(6x+1)}{(1+2x)^{61}(1-2x)^{41}}$$

A: Or write it this way too it's easier 
$P(x)=\frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}}$
$P(x)=(\frac{(1+x)^{10}}{(1-4x^2)^{4}(1+2x)^{2}})^{10}$
$P(x)=(\frac{(1+x)^{5}}{(1-4x^2)^{2}(1+2x)})^{20}=Q(x)^{20}$
Then apply $Q'(x)^n=nQ(x)^{n-1}Q'(x)$
A: 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}
A: HINT.-Take first derivative of $\dfrac{(1+x)^a}{(1-2x)^b(1+2x)^c}$ and replace the values of $a,b,c$ in your result. This way could be better for your calculations.
A: Alternatively:
$$[(1+x)^{100}(1-2x)^{-40}(1+2x)^{-60}]'=$$
$$100(1+x)^{99}(1-2x)^{-40}(1+2x)^{-60}+\\80(1+x)^{100}(1-2x)^{-41}(1+2x)^{-60}-\\120(1+x)^{100}(1-2x)^{-40}(1+2x)^{-61}=$$
$$20(1+x)^{99}(1-2x)^{-41}(1+2x)^{-61}\cdot \\ [5(1-2x)(1+2x)+4(1+x)(1+2x)-6(1+x)(1-2x)]=$$
$$\frac{20(1+x)^{99}}{(1-2x)^{41}(1+2x)^{61}}\cdot3(1+6x.)$$
