evaluate the derivative of the function.\ny = sec^{-1}(9 ln 3x)\n\\frac{dy}{dx}=\\square\n(use parentheses…

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(3x)), 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)

Differentiate (u = 9\ln(3x)) with respect to (x). Using the rule ((\ln(v))^\prime=\frac{v^\prime}{v}), where (v = 3x) and (v^\prime=3). So (u^\prime=\frac{9\times3}{3x}=\frac{9}{x}).

Step3: Combine the results

By the chain rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). Substitute (\frac{dy}{du}=\frac{1}{\vert9\ln(3x)\vert\sqrt{(9\ln(3x))^{2}-1}}) and (\frac{du}{dx}=\frac{9}{x}). Since (x>0) (for the domain of (\ln(3x)) when considering the real - valued function and assuming the principal branch of (\sec^{-1})), (\vert9\ln(3x)\vert = 9\vert\ln(3x)\vert) (and we can further note the domain conditions for the existence of the derivative based on the domain of (\sec^{-1}(t)) where (t = 9\ln(3x)) and (|t|\geq1)). (\frac{dy}{dx}=\frac{9}{x\vert9\ln(3x)\vert\sqrt{(9\ln(3x))^{2}-1}}=\frac{1}{x\vert\ln(3x)\vert\sqrt{81(\ln(3x))^{2}-1}})

Answer:

(\frac{1}{x\vert\ln(3x)\vert\sqrt{81(\ln(3x))^{2}-1}})