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

evaluate the derivative of the function.\n\n( y=sec ^{-1}(9 ln 7 x) )\n\nto 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\right)=\frac{1}{|u| sqrt{u^{2}-1}} \frac{d u}{d x} )\n\nuse the chain rule to find ( \frac{d u}{d x}=\frac{d}{d x}(9 ln 7 x) ).\n\n( \frac{d}{d x}(9 ln 7 x)=\frac{9}{x} )\n(simplify your answer.)\n\nreplace ( u ) and ( \frac{d u}{d x} ) with the terms found in previous steps in the formula for the derivative.\n\n( \frac{1}{|u| sqrt{u^{2}-1}} \frac{d u}{d x}=square )\n(use parentheses to clearly denote the argument of each function.)

evaluate the derivative of the function.\n\n( y=sec ^{-1}(9 ln 7 x) )\n\nto 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\right)=\frac{1}{|u| sqrt{u^{2}-1}} \frac{d u}{d x} )\n\nuse the chain rule to find ( \frac{d u}{d x}=\frac{d}{d x}(9 ln 7 x) ).\n\n( \frac{d}{d x}(9 ln 7 x)=\frac{9}{x} )\n(simplify your answer.)\n\nreplace ( u ) and ( \frac{d u}{d x} ) with the terms found in previous steps in the formula for the derivative.\n\n( \frac{1}{|u| sqrt{u^{2}-1}} \frac{d u}{d x}=square )\n(use parentheses to clearly denote the argument of each function.)

Answer

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}})