aufgabe 2a

$$f_1(x)=\frac{\sqrt{x}+x}{\sqrt{x}-x}$$

Ableitung §

$$\begin{array}{rl}{f}_1^{\prime }(x)& =\frac{{\left(\sqrt{x}+x\right)}^{\prime }\cdot \left(\sqrt{x}-x\right)-\left(\sqrt{x}+x\right)\cdot {\left(\sqrt{x}-x\right)}^{\prime }}{{\left(\sqrt{x}-x\right)}^2}\\ & =\frac{{\left(x^{\frac{1}{2}}+x\right)}^{\prime }\cdot \left(x^{\frac{1}{2}}-x\right)-\left(x^{\frac{1}{2}}+x\right)\cdot {\left(x^{\frac{1}{2}}-x\right)}^{\prime }}{{\left(\sqrt{x}-x\right)}^2}\\ & =\frac{\left(\frac{1}{2}{x}^{-\frac{1}{2}}+1\right)\cdot \left(x^{\frac{1}{2}}-x\right)-\left(x^{\frac{1}{2}}+x\right)\cdot \left(\frac{1}{2}{x}^{-\frac{1}{2}}-1\right)}{{\left(\sqrt{x}-x\right)}^2}\end{array}$$

Vereinfachen §

$$\begin{array}{rl}& =\frac{\left(\frac{1}{2}{x}^{-\frac{1}{2}}\cdot x^{\frac{1}{2}}+1\cdot x^{\frac{1}{2}}-\frac{1}{2}{x}^{-\frac{1}{2}}\cdot x-1\cdot x\right)-\left(x^{\frac{1}{2}}\cdot \frac{1}{2}{x}^{-\frac{1}{2}}+x\cdot \frac{1}{2}{x}^{-\frac{1}{2}}-x^{\frac{1}{2}}\cdot 1-x\cdot 1\right)}{{\left(\sqrt{x}-x\right)}^2}\\ & =\frac{\left(\frac{1}{2}{x}^0+x^{\frac{1}{2}}-\frac{1}{2}{x}^{\frac{1}{2}}-x\right)-\left(\frac{1}{2}{x}^0+\frac{1}{2}{x}^{\frac{1}{2}}-x^{\frac{1}{2}}-x\right)}{{\left(\sqrt{x}-x\right)}^2}\\ & =\frac{\frac{1}{2}+x^{\frac{1}{2}}-\frac{1}{2}{x}^{\frac{1}{2}}-x-\frac{1}{2}-\frac{1}{2}{x}^{\frac{1}{2}}+x^{\frac{1}{2}}+x}{{\left(\sqrt{x}-x\right)}^2}\\ & =\frac{\sqrt{x}}{{\left(\sqrt{x}-x\right)}^2}\end{array}$$

Beachten

• Quotientenregel benutzen
• $x^{\prime }=1$