91Ó°ÊÓ

To differentiate the given function, we can use the chain rule. The chain rule states that: \((g \circ h)'(x) = g'(h(x)) \cdot h'(x).\) In our case, \(g(u)=\ln u\) and \(h(x)=\sin x\). First, find the derivative of the outer function, \(g(u)\): \(g'(u)=\frac{1}{u}\). Next, find the derivative of the inner function, \(h(x)\): \(h'(x)=\cos x\). Now, apply the chain rule to find the derivative of the function: \(f^{\prime}(x)=g'(h(x)) \cdot h'(x) = \frac{1}{\sin x} \times \cos x\).

Now we'll evaluate the derivative \(f^{\prime}(x)\) at \(x=\frac{\pi}{6}\): \(f^{\prime}\left(\frac{\pi}{6}\right)=\frac{1}{\sin \left(\frac{\pi}{6}\right)} \times \cos \left(\frac{\pi}{6}\right)\). The sine and cosine of \(\frac{\pi}{6}\) are well-known: \(\sin \left(\frac{\pi}{6}\right)=\frac{1}{2}\), and \(\cos \left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\). Substitute these values into the expression: \(f^{\prime}\left(\frac{\pi}{6}\right)=\frac{1}{\frac{1}{2}} \times \frac{\sqrt{3}}{2} = 2 \times \frac{\sqrt{3}}{2}=\sqrt{3}\). It turns out that there is a mistake in the given derivative's expression. The corrected expression is \(f^{\prime}(x)=\frac{1}{\sin x} \times \cos x\), and the derivative for \(x=\frac{\pi}{6}\) is \(f^{\prime}\left(\frac{\pi}{6}\right)=\sqrt{3}\).