Question

evaluate the derivative of the function.

( y=sec ^{-1}(9 ln 7 x) )

( \frac{d y}{d x}= )
(use parentheses to clearly denote the argument of each function.)

Explanation:

Step1: Apply the chain rule

Let \(u = 9\ln(7x)\). The derivative of \(y=\sec^{-1}(u)\) with respect to \(u\) is \(\frac{1}{|u|\sqrt{u^{2}-1}}\).

Step2: Differentiate \(u = 9\ln(7x)\)

Using the chain - rule for \(u = 9\ln(7x)\), where the derivative of \(\ln(v)\) with respect to \(v\) is \(\frac{1}{v}\). Let \(v = 7x\), then \(\frac{du}{dx}=9\times\frac{1}{7x}\times7=\frac{9}{x}\).

Step3: Combine using the chain rule

By the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}\). Substituting \(y = \sec^{-1}(u)\) and \(u = 9\ln(7x)\) and \(\frac{du}{dx}=\frac{9}{x}\), we get \(\frac{dy}{dx}=\frac{1}{|9\ln(7x)|\sqrt{(9\ln(7x))^{2}-1}}\times\frac{9}{x}\).

Since \(9\ln(7x)\) is positive for \(x>\frac{1}{7}\) (assuming the domain where the function is well - defined), \(|9\ln(7x)| = 9\ln(7x)\).

Answer:

\(\frac{9}{x|9\ln(7x)|\sqrt{(9\ln(7x))^{2}-1}}\) (or \(\frac{1}{x\ln(7x)\sqrt{(9\ln(7x))^{2}-1}}\) when \(9\ln(7x)>0\))