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

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

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

Answer

Explanation:

Step1: Identify the inner function

Let (u = 9\ln(7x)).

Step2: Differentiate (u) with respect to (x)

Using the chain - rule for differentiation, (\frac{du}{dx}=\frac{9}{x}) (since (\frac{d}{dx}(\ln(ax))=\frac{a}{ax}=\frac{1}{x}) for (a = 7) and then multiply by 9).

Step3: Apply the formula for the derivative of (y=\sec^{-1}(u))

The formula is (\frac{dy}{dx}=\frac{1}{|u|\sqrt{u^{2}-1}}\frac{du}{dx}). Substitute (u = 9\ln(7x)) and (\frac{du}{dx}=\frac{9}{x}) into the formula. Since (9\ln(7x)>0) for (x > \frac{1}{7}) (assuming the domain where the function is well - defined), (|u|=u). (\frac{dy}{dx}=\frac{1}{9\ln(7x)\sqrt{(9\ln(7x))^{2}-1}}\cdot\frac{9}{x}) Simplify the expression: (\frac{dy}{dx}=\frac{1}{x\ln(7x)\sqrt{81(\ln(7x))^{2}-1}})

Answer:

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