Question

evaluate the derivative of the function.

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

to find ( \frac{d y}{d x} ), let ( u=9 ln (7 x) ) and use the derivative formula ( \frac{d}{d x}left(sec ^{-1} u
ight)=\frac{1}{|u| sqrt{u^{2}-1}} \frac{d u}{d x} )

use the chain rule to find ( \frac{d u}{d x}=\frac{d}{d x}(9 ln 7 x) ).

( \frac{d}{d x}(9 ln 7 x)=\frac{9}{x} )
(simplify your answer.)

replace ( u ) and ( \frac{d u}{d x} ) with the terms found in previous steps in the formula for the derivative.

( \frac{1}{|u| sqrt{u^{2}-1}} \frac{d u}{d x}=square )
(use parentheses to clearly denote the argument of each function.)

Explanation:

Step1: Substitute \(u\) and \(\frac{du}{dx}\)

Given \(u = 9\ln(7x)\) and \(\frac{du}{dx}=\frac{9}{x}\).
Substitute into \(\frac{1}{|u|\sqrt{u^{2}-1}}\frac{du}{dx}\).
We get \(\frac{1}{|9\ln(7x)|\sqrt{(9\ln(7x))^{2}-1}}\cdot\frac{9}{x}\).
Since \(9\ln(7x)\) is positive for \(x>\frac{1}{7}\) (domain - relevant for the inverse - secant function, and we can drop the absolute value for the non - negative case in the context of the derivative formula application here), the expression becomes \(\frac{9}{x\cdot9\ln(7x)\sqrt{81(\ln(7x))^{2}-1}}\).

Step2: Simplify the expression

Cancel out the common factor of \(9\) in the numerator and denominator.
The simplified expression is \(\frac{1}{x\ln(7x)\sqrt{81(\ln(7x))^{2}-1}}\).

Answer:

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