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:

This is a function inside a function inside a function. Start by naming the innermost part $v = \dfrac{3+2x}{3-2x}$, then work outward applying the chain rule for $\tan(\log v)$ one layer at a time. Before differentiating, try splitting $\log v$ using $\log(a/b) = \log a - \log b$ so you never need the 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 \(A = 12, B = 9, C = -4\).

Answer 2

Concept:

• $\log(a/b) = \log a - \log b$ turns the log of a quotient into a difference of two logs, so the argument never needs the quotient rule at all.
• Chain rule for tangent: $\dfrac{d}{dx}[\tan \theta] = \sec^2\theta \cdot \dfrac{d\theta}{dx}$.
• $\dfrac{d}{dx}[\log(\text{linear expression})] = \dfrac{\text{derivative of the expression}}{\text{the expression itself}}$, a direct one-line formula for each log term.

Step 1: Rewrite the inner function using the quotient rule for logarithms.
Let $\theta = \log\left(\dfrac{3+2x}{3-2x}\right) = \log(3+2x) - \log(3-2x)$, so $y = \tan\theta$.

Step 2: Differentiate each log term separately.
$\dfrac{d}{dx}[\log(3+2x)] = \dfrac{2}{3+2x}$ and $\dfrac{d}{dx}[\log(3-2x)] = \dfrac{-2}{3-2x}$.
So $\dfrac{d\theta}{dx} = \dfrac{2}{3+2x} - \left(\dfrac{-2}{3-2x}\right) = \dfrac{2}{3+2x} + \dfrac{2}{3-2x}$.

Step 3: Combine over a common denominator.
$\dfrac{d\theta}{dx} = \dfrac{2(3-2x) + 2(3+2x)}{(3+2x)(3-2x)} = \dfrac{12}{9 - 4x^2}$.

Step 4: Apply the chain rule for tangent and compare with the given form.
$\dfrac{dy}{dx} = \sec^2\theta \cdot \dfrac{d\theta}{dx} = \dfrac{12}{9-4x^2} \cdot \sec^2\left(\log\dfrac{3+2x}{3-2x}\right)$.
Comparing with $\dfrac{A}{B+Cx^2}\sec^2(\dots)$ gives $A = 12$, $B = 9$, $C = -4$.

Final Answer: $A = 12,\ B = 9,\ C = -4$.