Question

If \(y = f\left(\frac{3 + 2x}{3 - 2x}\right)\), where \(f(x) = \tan(\log x)\), and \(\frac{dy}{dx} = \frac{A}{B + Cx^2} \cdot \sec^2\left(\log \frac{3 + 2x}{3 - 2x}\right)\), then find \(A, B, C\).

Hint:

The derivative of \(\log(\frac{a+bx}{a-bx})\) is always \(\frac{2ab}{a^2 - b^2x^2}\). Memorizing this pattern helps in solving differentiation problems quickly without doing the full quotient rule!

Answer 1

Step 1: Understanding the Concept:
We are given a composite function \(y = f(g(x))\). We need to apply the Chain Rule: \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\).

Step 2: Key Formula or Approach:
1. \(\frac{d}{dx}[\tan(u)] = \sec^2(u) \frac{du}{dx}\).
2. \(\frac{d}{dx}[\log(v)] = \frac{1}{v} \frac{dv}{dx}\).
3. \(\frac{d}{dx}[\frac{3+2x}{3-2x}]\) using the Quotient Rule.

Step 3: Detailed Explanation:
Let \(u = \log\left(\frac{3+2x}{3-2x}\right)\). Then \(y = \tan(u)\).
\[ \frac{dy}{dx} = \sec^2(u) \cdot \frac{d}{dx} \left[ \log\left(\frac{3+2x}{3-2x}\right) \right] \] Let \(v = \frac{3+2x}{3-2x}\). Then \(\frac{d}{dx}(\log v) = \frac{1}{v} \frac{dv}{dx}\).
Calculate \(\frac{dv}{dx}\):
\[ \frac{dv}{dx} = \frac{(3-2x)(2) - (3+2x)(-2)}{(3-2x)^2} = \frac{6 - 4x + 6 + 4x}{(3-2x)^2} = \frac{12}{(3-2x)^2} \] Now, \(\frac{1}{v} \frac{dv}{dx} = \frac{3-2x}{3+2x} \cdot \frac{12}{(3-2x)^2} = \frac{12}{(3+2x)(3-2x)} = \frac{12}{9 - 4x^2}\).
Substituting back into the chain rule expression:
\[ \frac{dy}{dx} = \sec^2\left(\log \frac{3+2x}{3-2x}\right) \cdot \frac{12}{9 - 4x^2} \] Comparing with \(\frac{A}{B + Cx^2} \cdot \sec^2(\dots)\):
\(A = 12, B = 9, C = -4\).

Step 4: Final Answer:
The values are \(A = 12, B = 9, C = -4\).