Lösung Tangente in einem Kurvenpunkt III

\(h(x)=cos(\frac{\pi}{4}x)+1\)
\(h'(x)=\frac{\pi}{4}\cdot (-sin(\frac{\pi}{4}x))+1=-\frac{\pi}{4} sin(\frac{\pi}{4}x)\)
\(h'(6)=-\frac{\pi}{4}sin(\frac{\pi}{4}\cdot 6)=\frac{\pi}{4}\)