my function is: $y = \arcsin (2x+1)$ and I want to find its derivative.
My approach was to apply the chain rule:
$$y' = \frac{dg}{du} \frac{du}{dx}$$
with $g = \arcsin(u)$ and $u = 2x+1$.
$$g' = \frac{1}{\sqrt{1-u^2}}.$$
${u}' = 2$.
My solution therefore was $$\frac{1}{\sqrt{1-u^2}} \cdot 2 = \frac{2}{\sqrt{1-(2x+1)^2}}.$$
This seems to be wrong and the correct solution is given by: $\frac{1}{\sqrt{-x^{2}-x}}$
I know that implicit differentiation should be used for this particular problem, but I do not really understand why. I appreciate any help!
Your solution is correct; they just simplified it further: \begin{align*} \frac{2}{\sqrt{1 - (2x + 1)^2}} &= \frac{2}{\sqrt{1 - (4x^2 + 4x + 1)}} \\ &= \frac{2}{\sqrt{-4x^2 - 4x}} \\ &= \frac{2}{\sqrt{4(-x^2 - x)}} \\ &= \frac{2}{\sqrt{4}\sqrt{-x^2 - x}} \\ &= \frac{2}{2\sqrt{-x^2 - x}} \\ &= \frac{1}{\sqrt{-x^2 - x}} \\ \end{align*}