Question

help me solve this | 2 parts remaining
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)= ) (simplify your answer.)

Explanation:

Step1: Apply the constant multiple rule

The derivative of \(a\cdot f(x)\) is \(a\cdot f^{\prime}(x)\). Here \(a = 9\) and \(f(x)=\ln(7x)\). So \(\frac{d}{dx}(9\ln(7x))=9\frac{d}{dx}(\ln(7x))\).

Step2: Apply the chain - rule for \(\ln(u)\)

Let \(t = 7x\), and use the formula \(\frac{d}{dx}(\ln(t))=\frac{1}{t}\cdot\frac{dt}{dx}\). Since \(\frac{dt}{dx}=\frac{d}{dx}(7x) = 7\), then \(\frac{d}{dx}(\ln(7x))=\frac{1}{7x}\cdot7\).

Step3: Simplify the expression

\(9\frac{d}{dx}(\ln(7x))=9\cdot\frac{1}{7x}\cdot7\). The \(7\)s cancel out.

Answer:

\(\frac{9}{x}\)