First, let \(u = \sqrt{x}\), so \(f(x) = 2\cos^{-1}(u)\). Then, we have:
\(f'(x) = \frac{d}{dx}(2\cos^{-1}(u))\)
Now, we know that the derivative of \(\cos^{-1}(u)\) is \(-\frac{1}{\sqrt{1 - u^2}}\), so:
\(f'(x) = 2\cdot\left(-\frac{1}{\sqrt{1 -
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