Question

HELP PLEASE! no idea how to start this.

Answer 1

\[f(x)=\arcsin \sqrt{2x-1}\]

Answer 2

derivative? integral?

Answer 3

derivative, sorry.

Answer 4

you can always take the sin of both sides to arrive at the following:
\[y= \arcsin \sqrt{2x-1}\]
\[siny = \cancel{\sin(\arcsin)}\sqrt{2x-1}\]
\[siny = \sqrt{2x-1}\]
followdd by implicit differentiation:
\[\frac{d}{dx}siny= \frac{d}{dx}(2x-1)^{\frac{1}{2}}\]
\[cosy \frac{dy}{dx}=\frac{1}{2}(2x-1)^{-\frac{1}{2}}*(2)\]
\[\frac{dy}{dx}=\huge{\frac{1}{cosy*(2x-1)^{\frac{1}{2}}}}\]

Answer 5

remembering that you can solve for cosy based off of your siny term:
\[siny = \frac{\sqrt{2x-1}}{1}\]
where we've filled in the opposite and hypotenuse of the triangle (as per soh cah toa)
now have to solve cosy