Question

If \[ y=(1-x)(1+x^2)(1+x^4)\cdots(1+x^{2^n}), \] then \(\dfrac{dy}{dx}\) at \(x=0\) is equal to

• A. \(-1\)
• B. \(\dfrac{1}{(1+x)^2}\)
• C. \(\dfrac{x}{(1+x^2)}\)
• D. \(\dfrac{x}{(1-x)^2}\)

Hint:

When a function is a product of many terms, logarithmic differentiation simplifies the derivative. At \(x=0\), terms containing powers of \(x\) higher than one vanish.

Answer 1

The Correct Option is A

Step 1: Find the value of \(y\) at \(x=0\). \[ y(0)=(1-0)(1+0)(1+0)\cdots=1 \]
Step 2: Take logarithmic differentiation. \[ \ln y=\ln(1-x)+\ln(1+x^2)+\ln(1+x^4)+\cdots+\ln(1+x^{2^n}) \] Differentiate both sides: \[ \frac{1}{y}\frac{dy}{dx} = -\frac{1}{1-x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \cdots \]
Step 3: Evaluate at \(x=0\). At \(x=0\), \[ \frac{2x}{1+x^2}=\frac{4x^3}{1+x^4}=\cdots=0 \] Thus, \[ \left.\frac{1}{y}\frac{dy}{dx}\right|_{x=0}=-1 \] Since \(y(0)=1\), \[ \left.\frac{dy}{dx}\right|_{x=0}=-1 \]
Hence, \[ \boxed{-1} \]