Question

If \(y = f(\frac{3+2x}{3-2x})\), where \(f(x) = tan(logx)\) and \(\frac{dy}{dx} = (\frac{A}{B+Cx^2})\cdot sec^2(log(\frac{3+2x}{3-2x}))\), then the respective values of A, B and C are ......

• A. \(3,2,-4\)
• B. \(12,-9,4\)
• C. \(6,6,-2\)
• D. \(12,9,-4\)

Hint:

Use the chain rule on tan(log u), then simplify the derivative of log((3+2x)/(3-2x)) with the log difference rule.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The function is \(y=\tan\left(\log\frac{3+2x}{3-2x}\right)\). We use the chain rule: \(\frac{d}{dx}\tan u=\sec^2u\cdot u^{\prime}\).

Step 2: Differentiate the outer function:
\[ \frac{dy}{dx}=\sec^2\left(\log\frac{3+2x}{3-2x}\right)\cdot\frac{d}{dx}\left[\log\frac{3+2x}{3-2x}\right] \]

Step 3: Simplify the inner derivative:
\(\log\frac{3+2x}{3-2x}=\log(3+2x)-\log(3-2x)\). So
\[ \frac{2}{3+2x}+\frac{2}{3-2x}=\frac{2(3-2x)+2(3+2x)}{9-4x^2}=\frac{12}{9-4x^2} \]

Step 4: Compare with the given form:
The form \(\frac{A}{B+Cx^2}\) matches \(\frac{12}{9-4x^2}\), so \(A=12\), \(B=9\), \(C=-4\).

Step 5: Why the other options are wrong:
Option (B) \(12,-9,4\) gives \(\frac{12}{-9+4x^2}\), which has the wrong sign. Options (A) and (C) have the wrong numerator. Only \((12,9,-4)\) matches.

Final Answer:
The values are \(A=12\), \(B=9\), \(C=-4\), option (D). \[ \boxed{D} \]