Ableitung/Quotientenregel/1/Aufgabe/Lösung

Es ist

${}{\begin{aligned}f'(x)&={\frac {{\left(e^{x}-\cos x\right)}'{\left(x^{2}+5x-6\right)}-{\left(e^{x}-\cos x\right)}{\left(x^{2}+5x-6\right)}'}{{\left(x^{2}+5x-6\right)}^{2}}}\\&={\frac {{\left(e^{x}+\sin x\right)}{\left(x^{2}+5x-6\right)}-{\left(e^{x}-\cos x\right)}{\left(2x+5\right)}}{x^{4}+10x^{3}+13x^{2}-60x+36}}\\&={\frac {x^{2}e^{x}+5xe^{x}-6e^{x}+x^{2}\sin x+5x\sin x-6\sin x-2xe^{x}-5e^{x}+2x\cos x+5\cos x}{x^{4}+10x^{3}+13x^{2}-60x+36}}\\&={\frac {x^{2}e^{x}+3xe^{x}-11e^{x}+x^{2}\sin x+5x\sin x-6\sin x+2x\cos x+5\cos x}{x^{4}+10x^{3}+13x^{2}-60x+36}}.\end{aligned}}$

Zur gelösten Aufgabe