Cos Inverse Trig Derivative

[Jason76]

Given:

\(\displaystyle \dfrac{dy}{dx} \arccos u = \dfrac{-1}{\sqrt{1 - u^{2}}} (\dfrac{du}{dx})\)


\(\displaystyle y = \arccos(1 - 2x)\)

\(\displaystyle u = (1 - 2x)\)

\(\displaystyle du = -2\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{-1}{\sqrt{1 -(1 - 2x)^{2}}} (-2)\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{2}{\sqrt{1 - 1 + 4x - 4x^{2}}}\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{2}{\sqrt{4x -4x^{2}}}\) How does this line become the next line?

\(\displaystyle \dfrac{dy}{dx} = \pm \dfrac{1}{\sqrt{x - x^{2}}}\) Final Answer

 

[HallsofIvy]

Given:

\(\displaystyle \dfrac{dy}{dx} \arccos u = \dfrac{-1}{\sqrt{1 - u^{2}}} (\dfrac{du}{dx})\)


\(\displaystyle y = \arccos(1 - 2x)\)

\(\displaystyle u = (1 - 2x)\)

\(\displaystyle du = -2\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{-1}{\sqrt{1 -(1 - 2x)^{2}}} (-2)\)

\(\displaystyle \dfrac{dy}{dx} = \dfrac{2}{\sqrt{1 - 1 + 4x - 4x^{2}}}\)

\(\displaystyle \dfrac{dy}{dx} = \pm\dfrac{2}{\sqrt{4x -4x^{2}}}\) How does this line become the next line?

\(\displaystyle \dfrac{dy}{dx}= \pm\dfrac{2}{\sqrt{4(x- x^2)}}\)
\(\displaystyle \dfrac{dy}{dx}= \pm\dfrac{2}{\sqrt{4}\sqrt{x- x^2}}\)
\(\displaystyle \dfrac{dy}{dx}= \pm\dfrac{2}{2\sqrt{x- x^2}}\)

\(\displaystyle \dfrac{dy}{dx} = \pm \dfrac{1}{\sqrt{x - x^{2}}}\) Final Answer

(You should have the "\(\displaystyle \pm\)" on the very first square root.)