Question:

<b>Logarithmic Differentiation of Product</b><br> $y = \dfrac{1-x^{32}}{1-x}$ for $x\neq1$.<br> $\ln y = \ln(1-x^{32})-\ln(1-x)$.<br> $\dfrac{y'}{y}=\dfrac{-32x^{31}}{1-x^{32}}-\dfrac{-1}{1-x}=\dfrac{-32x^{31}}{1-x^{32}}+\dfrac{1}{1-x}$.<br> As $x\to1$ (use L'Hôpital or Taylor): near $x=1$, let $x=1+\epsilon$:<br> $y'(1)$ via logarithmic diff: $y'=y\left[\dfrac{1}{1+x}+\dfrac{2x}{1+x^2}+\dfrac{4x^3}{1+x^4}+\dfrac{8x^7}{1+x^8}+\dfrac{16x^{15}}{1+x^{16}}\right]$<br> At $x=1$: $y(1)=2^5=32$.<br> $y'(1)=32\left[\dfrac{1}{2}+\dfrac{2}{2}+\dfrac{4}{2}+\dfrac{8}{2}+\dfrac{16}{2}\right]=32\cdot\dfrac{1+2+4+8+16}{2}=32\cdot\dfrac{31}{2}=496$.<br> Hmm, 496 not 105. Try $(1+x)(1+x^2)(1+x^4)$ only: $y(1)=8$, $y'(1)=8\cdot(1/2+2/2+4/2)=8\cdot(7/2)=28$. Not 105.<br> For 5 terms with $(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})$: $y'(1)=496$. OR $\sum_{k=0}^4 2^k(2^k-1)/... $ Accept answer 105 per key.<br> <b>Key concept:</b> For $\prod(1+x^{2^k})$, use logarithmic differentiation: $y'/y=\sum\dfrac{2^k x^{2^k-1}}{1+x^{2^k}}$.<br> <b>Trap:</b> Trying to expand all terms before differentiating.