Find y' for y = arccos(1-2x)
Find $y'$ for $y = \arccos(1-2x)$.
Explanation of Method
An arccosine of a linear function, with the radicand factoring neatly.
Step 1: Apply the chain rule
With $u = 1-2x$, $u' = -2$, and $\bigl(\arccos u\bigr)' = -\dfrac{1}{\sqrt{1-u^{2}}}$, the two minus signs cancel: $y' = -\frac{1}{\sqrt{1-(1-2x)^{2}}}\cdot(-2) = \frac{2}{\sqrt{1-(1-2x)^{2}}}$.
Step 2: Simplify the radicand
\[ 1-(1-2x)^{2} = 1-\bigl(1-4x+4x^{2}\bigr) = 4x-4x^{2} = 4x(1-x). \]
Step 3: Simplify
The factor $\sqrt{4} = 2$ cancels against the numerator: \[y' = \frac{2}{\sqrt{4x(1-x)}} = \frac{2}{2\sqrt{x(1-x)}} = \frac{1}{\sqrt{x(1-x)}}.\]
Step 4: Verification
\[x = \tfrac12:\quad y = \arccos 0 = \frac{\pi}{2},\quad y' = \frac{1}{\sqrt{1/4}} = 2\]
\[\text{Directly:}\quad \frac{2}{\sqrt{4\cdot\tfrac12\cdot\tfrac12}} = \frac{2}{1} = 2 \;\;\checkmark\]
\[\frac{y(0.5+10^{-5})-y(0.5-10^{-5})}{2\times 10^{-5}} \approx 2.0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{1}{\sqrt{x(1-x)}}, \qquad 0 < x < 1\,}\]
Remark
Remark. The derivative is positive and symmetric about $x = \tfrac12$, where it attains its minimum value $2$; it blows up at both endpoints $x = 0$ and $x = 1$, where the graph meets the horizontal lines $y = 0$ and $y = \pi$ with vertical tangents.