TeX Quellcode:

\begin{array}{lclcrcl}\quad\mathbb{D}&=&\mathbb{R}\setminus_{\left\{10\right\}} \cr \cr f(x) &=& \dfrac{12 \sqrt{x}}{x - 10} \cr \cr \quad u(x) &=& 12 \sqrt{x} = 12x^{\frac{1}{2}} & \Rightarrow & u'(x) &=& 12\cdot\dfrac{1}{2}x^{-\frac{1}{2}} = \dfrac{6}{\sqrt{x}}\cr \quad v(x) &=&x - 10 & \Rightarrow & v'(x) &=& 1\cr \cr f'(x) &=& \dfrac{\dfrac{6}{\sqrt{x}}\cdot \left(x - 10\right)-12 \sqrt{x}\cdot 1}{\left(x - 10\right)^2} \cr \cr &=& \dfrac{\dfrac{6x - 60}{\sqrt{x}}-\dfrac{12x}{\sqrt{x}}}{\left(x - 10\right)^{2}} \cr \cr &=& \dfrac{\dfrac{-6 x - 60}{\sqrt{x}}}{\left(x - 10\right)^{2}} \cr \cr &=& -\dfrac{6 x + 60}{\sqrt{x}}\cdot\dfrac{1}{\left(x - 10\right)^{2}} \cr \cr &=& - \dfrac{6 x + 60}{\sqrt{x} \left(x - 10\right)^{2}} \end{array}