Question

evaluate the derivative of the function.
y = sec^{-1}(9 ln 3x)
\frac{dy}{dx}=\square
(use parentheses to clearly denote the argument of each function.)

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:

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