Find y' for y = x arccos x - sqrt(1-x²)
Find $y'$ for $y = x\arccos x - \sqrt{1-x^{2}}$.
Explanation of Method
Two awkward terms whose derivatives cancel exactly.
Step 1: Differentiate the product
With $u = x$, $u' = 1$, $v = \arccos x$, $v' = -\dfrac{1}{\sqrt{1-x^{2}}}$: $\frac{d}{dx}\bigl(x\arccos x\bigr) = \arccos x - \frac{x}{\sqrt{1-x^{2}}}$.
Step 2: Differentiate the radical
By the chain rule, and noting the leading minus sign: \[\frac{d}{dx}\!\left(-\sqrt{1-x^{2}}\right) = -\frac{-2x}{2\sqrt{1-x^{2}}} = \frac{x}{\sqrt{1-x^{2}}}.\]
Step 3: Combine
The two rational terms are exact negatives and cancel: $y' = \arccos x - \frac{x}{\sqrt{1-x^{2}}} + \frac{x}{\sqrt{1-x^{2}}} = \arccos x$.
Step 4: Verification
\[x = 0:\quad y = 0 - 1 = -1,\quad y' = \arccos 0 = \frac{\pi}{2} \approx 1.5708\]
\[\frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 1.5708 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \arccos x, \qquad |x| < 1\,}\]
Remark
Remark. Read backwards, this is exactly how $\displaystyle\int\arccos x\,dx = x\arccos x - \sqrt{1-x^{2}} + C$ is obtained by parts. Since $\arccos x \gt 0$ on $(-1,1)$, the function is strictly increasing there.