Question

evaluate the derivative of the function.

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

$\frac{dy}{dx}=\square$
(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 \(u = 9\ln(7x)\), 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:

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