Question

If \(y = (\frac{ax+b}{cx+d})\), then \(2\frac{dy}{dx}\cdot \frac{d^3y}{dx^3}\) is equal to

• A. \((\frac{d^2y}{dx^2})^2\)
• B. \(3\frac{d^2y}{dx^2}\)
• C. \(3(\frac{d^2y}{dx^2})^2\)
• D. \(3\frac{d^2x}{dy^2}\)

Hint:

Write y as a constant plus K/(cx+d) and differentiate three times.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A Mobius type function \(y=\dfrac{ax+b}{cx+d}\) can be written as a constant plus \(K(cx+d)^{-1}\).

Step 2: Rewrite:
\(y=\dfrac ac+\dfrac{bc-ad}{c}\cdot\dfrac1{cx+d}\). Put \(K=\dfrac{bc-ad}{c}\). Then \(y=\dfrac ac+K(cx+d)^{-1}\).

Step 3: Derivatives:
\(y'=-Kc(cx+d)^{-2}\)
\(y''=2Kc^2(cx+d)^{-3}\)
\(y'''=-6Kc^3(cx+d)^{-4}\)

Step 4: Form the products:
\(2y'y'''=2\cdot(-Kc)(-6Kc^3)(cx+d)^{-6}=12K^2c^4(cx+d)^{-6}\).
\((y'')^2=4K^2c^4(cx+d)^{-6}\).

Step 5: Compare:
\(2y'y'''=3(y'')^2\). Option (C). Option (A) is smaller by a factor 3, and (B) mixes in a linear term.

Final Answer:
2 y' y''' = 3 (y'')^2. \[ \boxed{3\left(\frac{d^2y}{dx^2}\right)^2} \]