int cos ^{-1}left(sqrt{frac{x}{a+x}}right) d | vedclass

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(D) Given that $\int \cos ^{-1} \sqrt{\frac{x}{a+x}} d x = f(x) + C$. By the Fundamental Theorem of Calculus,the derivative of the integral is the int

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(D) Given that $\int \cos ^{-1} \sqrt{\frac{x}{a+x}} d x = f(x) + C$. By the Fundamental Theorem of Calculus,the derivative of the integral is the integrand: $f^{\prime}(x) = \frac{d}{dx} \int \cos ^{-1} \sqrt{\frac{x}{a+x}} d x = \cos ^{-1} \sqrt{\frac{x}{a+x}}$. Now,we need to find $f^{\prime}(a)$. Substitute $x = a$ into the expression for $f^{\prime}(x)$: $f^{\prime}(a) = \cos ^{-1} \sqrt{\frac{a}{a+a}} = \cos ^{-1} \sqrt{\frac{a}{2a}}$. Simplifying the fraction inside the square root: $f^{\prime}(a) = \cos ^{-1} \sqrt{\frac{1}{2}} = \cos ^{-1} \left(\frac{1}{\sqrt{2}}\right)$. Since $\cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$,we have: $f^{\prime}(a) = \frac{\pi}{4}$.

$\int \cos ^{-1}\left(\sqrt{\frac{x}{a+x}}\right) d x=f(x)+C \Rightarrow f^{\prime}(a)=$

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