help me solve this | 2 parts remaining\nevaluate the derivative of the function.\n( y=sec ^{-1}(9 ln 7 x)…

help me solve this | 2 parts remaining\nevaluate the derivative of the function.\n( y=sec ^{-1}(9 ln 7 x) )\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} ).\nuse the chain rule to find ( \frac{d u}{d x}=\frac{d}{d x}(9 ln 7 x) ).\n( \frac{d}{d x}(9 ln 7 x)= ) (simplify your answer.)

help me solve this | 2 parts remaining\nevaluate the derivative of the function.\n( y=sec ^{-1}(9 ln 7 x) )\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} ).\nuse the chain rule to find ( \frac{d u}{d x}=\frac{d}{d x}(9 ln 7 x) ).\n( \frac{d}{d x}(9 ln 7 x)= ) (simplify your answer.)

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})