Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{120}D^{10}\: \frac{1}{x+2}\; =$	A. \(\boldsymbol{10}\)
B. \(\boldsymbol{\frac {10\,!} {(x\,+\,2)^{10}} }\)
C. \(\boldsymbol{-\,\frac {10\,!} {(x\,+\,2)^{10}} }\)
D. \(\boldsymbol{\frac {10\,!} {(x\,+\,2)^{11}} } \)
E. \(\boldsymbol{-\,\frac {10\,!} {(x\,+\,2)^{11}} }\)

$\begin{aligned}&\mathbf{D}^{10}\frac{1}{x+2}=\mathbf{D}^9\left(-\frac{1}{(x+2)^2}\right)=\mathbf{D}^8\left(\frac{2.(x+2)}{(x+2)^4}\right)\\&=2.\mathbf{D}^8\left(\frac{1}{(x+2)^3}\right)=2.\mathbf{D}^7\left(-\frac{3(x+2)^2}{(x+2)^6}\right)=-2.3.\mathbf{D}^7\left(\frac{1}{(x+2)^4}\right)\\&=-2.3.\mathbf{D}^6\left(\frac{4(x+2)^3}{(x+2)^8}\right)=+2.3.4.\mathbf{D}^6\left(\frac{1}{(x+2)^5}\right)\\&=\cdots=2.3.4.5.6.7.8.9.10.\frac{1}{(x+2)^{11}}=\frac{10!}{(x+2)^{11}}\end{aligned}$