The domain of the derivative of the function | vedclass

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(C) To find the domain of the derivative $f'(x)$, we first determine the domain of the function $f(x) = \operatorname{Cos}^{-1}(2x - 5) - \operatornam

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$(2, 3)$

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(C) To find the domain of the derivative $f'(x)$, we first determine the domain of the function $f(x) = \operatorname{Cos}^{-1}(2x - 5) - \operatorname{Sin}^{-1}(x - 2)$.For $\operatorname{Cos}^{-1}(2x - 5)$ to be defined, $-1 \le 2x - 5 \le 1$, which implies $4 \le 2x \le 6$, so $2 \le x \le 3$.For $\operatorname{Sin}^{-1}(x - 2)$ to be defined, $-1 \le x - 2 \le 1$, which implies $1 \le x \le 3$.The intersection of these intervals is $[2, 3]$.Now, we find the derivative $f'(x) = -\frac{1}{\sqrt{1 - (2x - 5)^2}} \cdot 2 - \frac{1}{\sqrt{1 - (x - 2)^2}} \cdot 1$.The derivative is defined where the expressions inside the square roots are strictly positive.For $\operatorname{Cos}^{-1}(2x - 5)$, the derivative is undefined at $2x - 5 = \pm 1$, i.e.,$x = 2$ and $x = 3$.For $\operatorname{Sin}^{-1}(x - 2)$, the derivative is undefined at $x - 2 = \pm 1$, i.e.,$x = 1$ and $x = 3$.Thus, the derivative is defined in the open interval $(2, 3)$ excluding the points where the denominators become zero.Therefore, the domain of the derivative is $(2, 3)$.

The domain of the derivative of the function $f(x) = \operatorname{Cos}^{-1}(2x - 5) - \operatorname{Sin}^{-1}(x - 2)$ is

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