Kettenregel/R/Rationale Funktionen/Bestätige/Aufgabe/Lösung

1. Es ist

   ${}{\begin{aligned}h(x)&=g(f(x))\\&={\frac {f(x)+4}{f(x)^{2}-5}}\\&={\frac {{\frac {x^{2}-3}{x+2}}+4}{\left({\frac {x^{2}-3}{x+2}}\right)^{2}-5}}\\&={\frac {(x^{2}-3)(x+2)+4(x+2)^{2}}{(x^{2}-3)^{2}-5(x+2)^{2}}}\\&={\frac {x^{3}+2x^{2}-3x-6+4x^{2}+16x+16}{x^{4}-6x^{2}+9-5x^{2}-20x-20}}\\&={\frac {x^{3}+6x^{2}+13x+10}{x^{4}-11x^{2}-20x-11}}.\end{aligned}}$

2. Die Ableitung von ${}h$ ist nach der Quotientenregel gleich

   ${}{\begin{aligned}h'(x)&={\frac {(x^{3}+6x^{2}+13x+10)'(x^{4}-11x^{2}-20x-11)-(x^{3}+6x^{2}+13x+10)(x^{4}-11x^{2}-20x-11)'}{(x^{4}-11x^{2}-20x-11)^{2}}}\\&={\frac {(3x^{2}+12x+13)(x^{4}-11x^{2}-20x-11)-(x^{3}+6x^{2}+13x+10)(4x^{3}-22x-20)}{(x^{4}-11x^{2}-20x-11)^{2}}}\\&={\frac {3x^{6}+12x^{5}-20x^{4}-192x^{3}-416x^{2}-392x-143-4x^{6}-24x^{5}-30x^{4}+112x^{3}+406x^{2}+480x+200}{x^{8}+121x^{4}+400x^{2}+121-22x^{6}-40x^{5}-22x^{4}+440x^{3}+242x^{2}+440x}}\\&={\frac {-x^{6}-12x^{5}-50x^{4}-80x^{3}-10x^{2}+88x+57}{x^{8}-22x^{6}-40x^{5}+99x^{4}+440x^{3}+642x^{2}+440x+121}}.\end{aligned}}$

3. Es ist

   ${}f'(x)={\frac {2x(x+2)-(x^{2}-3)}{(x+2)^{2}}}={\frac {x^{2}+4x+3}{x^{2}+4x+4}}\,$

   und

   ${}g'(y)={\frac {y^{2}-5-(y+4)2y}{(y^{2}-5)^{2}}}={\frac {-y^{2}-8y-5}{y^{4}-10y^{2}+25}}\,.$

   Gemäß der Kettenregel ist somit

   ${}{\begin{aligned}h'(x)&=g'(f(x))f'(x)\\&={\frac {-\left({\frac {x^{2}-3}{x+2}}\right)^{2}-8\left({\frac {x^{2}-3}{x+2}}\right)-5}{\left({\frac {x^{2}-3}{x+2}}\right)^{4}-10\left({\frac {x^{2}-3}{x+2}}\right)^{2}+25}}\cdot {\frac {x^{2}+4x+3}{x^{2}+4x+4}}\\&={\frac {-\left({\frac {x^{2}-3}{x+2}}\right)^{2}-8\left({\frac {x^{2}-3}{x+2}}\right)-5}{\left({\frac {x^{2}-3}{x+2}}\right)^{4}-10\left({\frac {x^{2}-3}{x+2}}\right)^{2}+25}}\cdot {\frac {x^{2}+4x+3}{(x+2)^{2}}}\cdot {\frac {(x+2)^{2}}{(x+2)^{2}}}\\&={\frac {-(x^{2}-3)^{2}-8(x^{2}-3)(x+2)-5(x+2)^{2}}{(x^{2}-3)^{4}-10(x^{2}-3)^{2}(x+2)^{2}+25(x+2)^{4}}}\cdot {\left(x^{2}+4x+3\right)}\\&={\frac {-x^{4}-8x^{3}-15x^{2}+4x+19}{x^{8}-22x^{6}-40x^{5}+99x^{4}+440x^{3}+642x^{2}+440x+19}}\cdot {\left(x^{2}+4x+3\right)}\\&={\frac {-x^{6}-12x^{5}-50x^{4}-80x^{3}-10x^{2}+88x+57}{x^{8}-22x^{6}-40x^{5}+99x^{4}+440x^{3}+642x^{2}+440x+121}}\end{aligned}}$

Zur gelösten Aufgabe