$$y=(\sin(x))^{(\cos(x)-1)}$$
We multiply by $\ln$
$$\ln(y)=(\cos(x)-1)\cdot \ln(\sin(x))$$
$$\frac{y'}y=(\cos(x)-1)'\cdot \ln(\sin(x)) + (\cos(x)-1)\cdot \ln(\sin(x))'$$
$$\frac{y'}y=-\sin(x)\cdot \ln(\sin(x)) + (\cos(x)-1)\cdot \dfrac{\cos(x)}{\sin(x)}$$
$$y'=y\cdot\bigg[(\cos(x)-1)\cdot \dfrac{\cos(x)}{\sin(x)} - (\sin(x))\cdot \ln(\sin(x))\bigg]$$
I am not sure if this is correct. If not help.