Find y' for y = ((1-x²)e^(3x-1)cos x)/((arccos x)³)

Find $y'$ for $y = \dfrac{\bigl(1-x^{2}\bigr)e^{3x-1}\cos x}{(\arccos x)^{3}}$.

Step 1: Take the natural logarithm

\[ \ln y = \ln\bigl(1-x^{2}\bigr) + (3x-1) + \ln\cos x - 3\ln\bigl(\arccos x\bigr). \]

Step 2: Differentiate term by term

The first three are routine; the last needs $\bigl(\arccos x\bigr)' = -\dfrac{1}{\sqrt{1-x^{2}}}$, and the two minus signs combine into a plus: \[\frac{d}{dx}\Bigl[-3\ln\arccos x\Bigr] = -3\cdot\frac{1}{\arccos x}\cdot\left(-\frac{1}{\sqrt{1-x^{2}}}\right) = \frac{3}{\arccos x\,\sqrt{1-x^{2}}}.\]

Altogether, using $\bigl(\ln\cos x\bigr)' = -\tan x$ from Problem 120: $\frac{y'}{y} = \frac{-2x}{1-x^{2}} + 3 - \tan x + \frac{3}{\arccos x\,\sqrt{1-x^{2}}}$.

Step 3: Verification

\[x = 0:\quad \arccos 0 = \tfrac{\pi}{2} \;\Longrightarrow\; y = \frac{1\cdot e^{-1}\cdot 1}{(\pi/2)^{3}} \approx 0.094917\]

\[\text{bracket} = 0 + 3 - 0 + \frac{3}{(\pi/2)\cdot 1} \approx 4.90986 \;\Longrightarrow\; y' \approx 0.46603\]

\[\frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 0.46603 \;\;\checkmark\]

Step 4: Domain

Three conditions: $1-x^{2} \gt 0$ for the first logarithm, $\arccos x \neq 0$ (so $x \neq 1$), and $\cos x \gt 0$ — automatic on $|x| \lt 1$ since $1 \lt \tfrac{\pi}{2}$. So the formula holds on $(-1,1)$.