evaluate the derivative of the function.\n\n$y = \\sec^{-1}(9\\ln 7x)$\n\n$\\frac{dy}{dx}=\\square$\n(use…

Explanation:

Step1: Apply the chain rule

The chain rule states that if (y = f(g(x))), then (y^\prime=f^\prime(g(x))\cdot g^\prime(x)). Let (u = 9\ln(7x)), so (y=\sec^{- 1}(u)). The derivative of (y = \sec^{-1}(u)) with respect to (u) is (\frac{dy}{du}=\frac{1}{\vert u\vert\sqrt{u^{2}-1}}).

Step2: Find the derivative of (u)

Now, find the derivative of (u = 9\ln(7x)). Using the rule ((\ln(v))^\prime=\frac{v^\prime}{v}), where (v = 7x) and (v^\prime=7). So (u^\prime=\frac{9\times7}{7x}=\frac{9}{x}).

Step3: Combine using the chain rule

By the chain rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). Substituting (u = 9\ln(7x)) and (u^\prime=\frac{9}{x}) into (\frac{dy}{dx}), we get (\frac{dy}{dx}=\frac{1}{\vert9\ln(7x)\vert\sqrt{(9\ln(7x))^{2}-1}}\cdot\frac{9}{x}). Since (x>0) (for the domain of (\ln(7x)) where (7x>0) i.e., (x > 0)) and (9\ln(7x)) is positive for (x> \frac{1}{7}) (when considering the domain of (\sec^{-1}(u)) where (|u|\geq1)), we can drop the absolute - value (assuming (x) is in the domain where (9\ln(7x)>0)). So (\frac{dy}{dx}=\frac{9}{x\sqrt{81(\ln(7x))^{2}-1}\cdot9\ln(7x)}=\frac{1}{x\ln(7x)\sqrt{81(\ln(7x))^{2}-1}}).

Answer:

(\frac{1}{x\ln(7x)\sqrt{81(\ln(7x))^{2}-1}})