If y=(1+x)(1+x^2)(1+x^4) dots (1+x^{2^n}), th | vedclass

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(C) Given $y = (1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$.Taking the natural logarithm on both sides:$\ln y = \ln(1+x) + \ln(1+x^2) + \ln(1+x^4) + \dots +

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(C) Given $y = (1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$.Taking the natural logarithm on both sides:$\ln y = \ln(1+x) + \ln(1+x^2) + \ln(1+x^4) + \dots + \ln(1+x^{2^n})$.Differentiating both sides with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \dots + \frac{2^n x^{2^n-1}}{1+x^{2^n}}$.Thus, $\frac{dy}{dx} = y \left[ \frac{1}{1+x} + \frac{2x}{1+x^2} + \dots + \frac{2^n x^{2^n-1}}{1+x^{2^n}} \right]$.At $x=0$, $y = (1+0)(1+0) \dots (1+0) = 1$.Substituting $x=0$ into the derivative expression:$\left(\frac{dy}{dx}\right)_{x=0} = 1 \cdot \left[ \frac{1}{1+0} + 0 + 0 + \dots + 0 \right] = 1 \cdot 1 = 1$.

If $y=(1+x)(1+x^2)(1+x^4) \dots (1+x^{2^n})$, then the value of $\left(\frac{dy}{dx}\right)_{x=0}$ is

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