Cześć!
[tex]f(x)=\mathrm{arctg(\frac{cosx-sin^2x}{\sqrt{ln(x+cosx)}})}[/tex]
Skorzystamy z własności pochodnej funkcji złożonej:
Niech [tex]a=\frac{\cos{x}-\sin^2{x}}{\sqrt{\ln{x+\cos{x}}}}[/tex], wówczas [tex]\mathrm{arctg(\frac{cosx-sin^2x}{\sqrt{ln(x+cosx)}})} = \mathrm{arctg}(a)[/tex]:
[tex]\frac{d}{dx}f(x) =\frac{d}{dx}a \cdot \frac{d}{dx}\mathrm{arctg}(a)[/tex]
1. Pochodna licznika
[tex]\mathr\frac{d}{dx}(\cos{x}-\sin^2{x})= \frac{d}{dx}(\cos{x}) - \frac{d}{dx}(\sin^2{x}) = -\sin{x} -\cos{x}\cdot 2\sin{x}=\\\\=-\sin{x}-2\sin{x}\cos{x} = -\sin{x}-\sin{2x}[/tex]
2. Pochodna mianownika
[tex]p = x+\cos{x}\\\\q = \ln(p)\\\\r=\sqrt{q}[/tex]
[tex]\frac{d}{dx}(\sqrt{\ln(x+\cos{x}}) = p' \cdot q' \cdot r' = (1-\sin{x})\cdot\frac{1}{x+\cos{x}} \cdot \frac{1}{2\sqrt{\ln(x+\cos{x}})} =\\\\= \frac{1-sinx}{(x+\cos{x})2\sqrt{\ln(x+\cos{x}})}[/tex]
Wówczas:
[tex]a' = \dfrac{\frac{d}{dx}(\cos{x}-\sin^2{x})\cdot \sqrt{\ln(x+\cos{x})} - (\cos{x}-\sin^2{x})\frac{d}{dx}(\sqrt{\ln(x+\cos{x})})}{\ln(x+\cos{x})}\\\\\\a' = \dfrac{(-\sin{x}-\sin{2x})\cdot \sqrt{\ln(x+\cos{x})} - (\cos{x}-\sin^2{x})\cdot \frac{1-\sin{x}}{(x+\cos{x})2\sqrt{\ln(x+\cos{x}})}}{\ln(x+\cos{x})}[/tex]
Korzystając z informacji, że [tex](\mathrm{arctg}(x))' = \frac{1}{1+x^2}[/tex]:
[tex]\frac{d}{dx}(\mathrm{arctg(\frac{cosx-sin^2x}{\sqrt{ln(x+cosx)}})} =\dfrac{1}{1+(\frac{\cos{x}-\sin^2{x}}{\sqrt{\ln{x+\cos{x}}}})^2} \cdot \\\\\\\dfrac{(-\sin{x}-\sin{2x})\cdot \sqrt{\ln(x+\cos{x})} - (\cos{x}-\sin^2{x})\cdot \frac{1-\sin{x}}{(x+\cos{x})2\sqrt{\ln(x+\cos{x}})}}{\ln(x+\cos{x})}[/tex]
Pozdrawiam!
