The second order derivative of cos^{-1}(4x^3 | vedclass

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(A) Let $u = \cos^{-1}(4x^3 - 3x)$ and $v = \cos^{-1}(2x^2 - 1)$.Given $\frac{1}{2} < x < 1$, we can substitute $x = \cos 	heta$, where $0 < 	heta <

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(A) Let $u = \cos^{-1}(4x^3 - 3x)$ and $v = \cos^{-1}(2x^2 - 1)$.Given $\frac{1}{2} Then $u = \cos^{-1}(\cos 3	heta) = 3	heta$ and $v = \cos^{-1}(\cos 2	heta) = 2	heta$.We need to find $\frac{d^2u}{dv^2}$.Since $u = 3	heta$ and $v = 2	heta$, we have $	heta = \frac{v}{2}$.Substituting $	heta$ in $u$, we get $u = 3(\frac{v}{2}) = \frac{3}{2}v$.Now, $\frac{du}{dv} = \frac{3}{2}$.Therefore, the second order derivative $\frac{d^2u}{dv^2} = \frac{d}{dv}(\frac{3}{2}) = 0$.

The second order derivative of $\cos^{-1}(4x^3 - 3x)$ with respect to $\cos^{-1}(2x^2 - 1)$, where $\frac{1}{2} < x < 1$ is

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