Question

calculus:find the derivative:

y=(x-1)*(squareroot of 2x-x^2)+arcsin(x-1)

Answer 1

first lets look at this part and we will gall it g:
\[g=\sin^{-1}(x-1) => \sin(g)=x-1\]

so anyways we want to find g' using implicit differentiation
\[\cos(g) g'=1 => g'=\frac{1}{\cos(g)}\]
we need this in terms of so referred to the triangle
\[\cos(g)=\frac{adjacent side of g}{hyp}=\frac{\sqrt{1-(x-1)^2}}{1}=\sqrt{1-(x^2-2x+1)}\]
so we have
\[g'=\frac{1}{\sqrt{1-x^2+2x-1}}=\frac{1}{\sqrt{2x-x^2}}\]