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

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))