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 Question:
We are given a composite function \( y = \tan\left(\log\left(\frac{3+2x}{3-2x}\right)\right) \). We need to differentiate it using the chain rule and compare the result with the given form to find the constants.
Step 2: Key Formula or Approach:
Chain Rule: \( \frac{d}{dx}[f(g(h(x)))] = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \).
Logarithmic property: \( \frac{d}{dx}(\log u) = \frac{1}{u} \frac{du}{dx} \).
Step 3: Detailed Explanation:
Let \( u = \frac{3+2x}{3-2x} \).
First, find \( \frac{d}{dx} (\log u) \):
\[ \frac{d}{dx} \left( \log \frac{3+2x}{3-2x} \right) = \frac{d}{dx} [\log(3+2x) - \log(3-2x)] \]
\[ = \frac{2}{3+2x} - \frac{-2}{3-2x} \]
\[ = \frac{2(3-2x) + 2(3+2x)}{(3+2x)(3-2x)} = \frac{6-4x+6+4x}{9-4x^2} = \frac{12}{9-4x^2} \]
Now, \( y = \tan(\log u) \).
\[ \frac{dy}{dx} = \sec^2(\log u) \cdot \frac{d}{dx}(\log u) \]
\[ \frac{dy}{dx} = \sec^2\left(\log \frac{3+2x}{3-2x}\right) \cdot \frac{12}{9-4x^2} \]
Comparing this with the given form \( \frac{A}{B + Cx^2} \sec^2(\dots) \):
\( A = 12 \), \( B = 9 \), \( C = -4 \).
Step 4: Final Answer:
The values are \( A=12, B=9, C=-4 \).