The derivative of the function f(x) = cos^{-1 | vedclass

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(A) Let $\cos \alpha = \frac{2}{\sqrt{13}}$ and $\sin \alpha = \frac{3}{\sqrt{13}}$. Then $	an \alpha = \frac{3}{2}$.The expression becomes:$f(x) = \

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(A) Let $\cos \alpha = \frac{2}{\sqrt{13}}$ and $\sin \alpha = \frac{3}{\sqrt{13}}$. Then $	an \alpha = \frac{3}{2}$.The expression becomes:$f(x) = \cos^{-1} (\cos \alpha \cos x - \sin \alpha \sin x) + \sin^{-1} (\cos \alpha \cos x + \sin \alpha \sin x)$Using trigonometric identities $\cos(A+B) = \cos A \cos B - \sin A \sin B$ and $\cos(A-B) = \cos A \cos B + \sin A \sin B$:$f(x) = \cos^{-1} (\cos(x + \alpha)) + \sin^{-1} (\cos(x - \alpha))$Since $\sin^{-1}(\cos 	heta) = \frac{\pi}{2} - \cos^{-1}(\cos 	heta) = \frac{\pi}{2} - 	heta$ (for appropriate range):$f(x) = (x + \alpha) + \left( \frac{\pi}{2} - (x - \alpha) ight)$$f(x) = x + \alpha + \frac{\pi}{2} - x + \alpha = \frac{\pi}{2} + 2\alpha$Since $\alpha$ is a constant,the derivative $f'(x) = \frac{d}{dx} (\frac{\pi}{2} + 2\alpha) = 0$.Thus,the derivative at $x = \frac{3}{4}$ is $0$.

The derivative of the function $f(x) = \cos^{-1} \left\{ \frac{1}{\sqrt{13}} (2\cos x - 3\sin x) \right\} + \sin^{-1} \left\{ \frac{1}{\sqrt{13}} (2\cos x + 3\sin x) \right\}$ with respect to $x$ at $x = \frac{3}{4}$ is:

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