evaluate the derivative of the following function.\nf(z)=8\\cot^{-1}\\sqrt{7z}\nf(z)=\\square

Explanation:

Step1: Recall the derivative formula for (y = \cot^{-1}(u))

The derivative of (y=\cot^{-1}(u)) with respect to (x) is (\frac{dy}{dx}=-\frac{1}{1 + u^{2}}\cdot\frac{du}{dx}) (by the chain rule). Here (u = \sqrt{7z}), and (f(z)=8\cot^{-1}(\sqrt{7z})).

Step2: Find the derivative of (u=\sqrt{7z})

First, rewrite (u=\sqrt{7z}=(7z)^{\frac{1}{2}}). Using the power rule (\frac{d}{dz}(az^{n})=anz^{n - 1}) (where (a = 7) and (n=\frac{1}{2})), we get (\frac{du}{dz}=\frac{7}{2}(7z)^{-\frac{1}{2}}=\frac{\sqrt{7}}{2\sqrt{z}}).

Step3: Apply the chain - rule to (f(z))

Since (f(z)=8\cot^{-1}(u)) with (u = \sqrt{7z}), by the chain rule (f^{\prime}(z)=8\times\left(-\frac{1}{1 + u^{2}}\right)\times\frac{du}{dz}). Substitute (u = \sqrt{7z}) into (-\frac{1}{1 + u^{2}}), we have (-\frac{1}{1+7z}). Then (f^{\prime}(z)=8\times\left(-\frac{1}{1 + 7z}\right)\times\frac{\sqrt{7}}{2\sqrt{z}}).

Step4: Simplify the expression

[ \begin{align*} f^{\prime}(z)&=8\times\left(-\frac{1}{1 + 7z}\right)\times\frac{\sqrt{7}}{2\sqrt{z}}\ &=-\frac{4\sqrt{7}}{\sqrt{z}(1 + 7z)} \end{align*} ]

Answer:

(-\frac{4\sqrt{7}}{\sqrt{z}(1 + 7z)})