If a b 0 and x is acute,then frac{d}{dx} | vedclass

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(B) Let $y = \cos^{-1} \left( \frac{b - a \cos x}{a - b \cos x} ight)$.Using the substitution $\cos 	heta = \frac{b - a \cos x}{a - b \cos x}$,we k

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$\frac{-\sqrt{a^2 - b^2}}{a - b \cos x}$

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(B) Let $y = \cos^{-1} \left( \frac{b - a \cos x}{a - b \cos x} ight)$.Using the substitution $\cos 	heta = \frac{b - a \cos x}{a - b \cos x}$,we know that $\cos y = \frac{b - a \cos x}{a - b \cos x}$.By applying the component-dividendo rule or trigonometric identities for $	an(y/2)$,we find that $	an \left( \frac{y}{2} ight) = \sqrt{\frac{1 - \cos y}{1 + \cos y}}$.Substituting $\cos y = \frac{b - a \cos x}{a - b \cos x}$:$1 - \cos y = 1 - \frac{b - a \cos x}{a - b \cos x} = \frac{a - b \cos x - b + a \cos x}{a - b \cos x} = \frac{(a - b)(1 + \cos x)}{a - b \cos x}$.$1 + \cos y = 1 + \frac{b - a \cos x}{a - b \cos x} = \frac{a - b \cos x + b - a \cos x}{a - b \cos x} = \frac{(a + b)(1 - \cos x)}{a - b \cos x}$.Thus,$	an^2 \left( \frac{y}{2} ight) = \frac{(a - b)(1 + \cos x)}{(a + b)(1 - \cos x)} = \frac{a - b}{a + b} \cot^2 \left( \frac{x}{2} ight)$.Taking the square root,$y = 2 	an^{-1} \left( \sqrt{\frac{a - b}{a + b}} \cot \frac{x}{2} ight)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 2 \cdot \frac{1}{1 + \frac{a - b}{a + b} \cot^2 \frac{x}{2}} \cdot \sqrt{\frac{a - b}{a + b}} \cdot \left( -\frac{1}{2} \csc^2 \frac{x}{2} ight)$.Simplifying this expression yields $\frac{dy}{dx} = \frac{-\sqrt{a^2 - b^2}}{a - b \cos x}$.

If $a > b > 0$ and $x$ is acute,then $\frac{d}{dx} \left[ \cos^{-1} \left( \frac{b - a \cos x}{a - b \cos x} \right) \right] = $

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