If $\mathrm{f}(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)$, then $f^{\prime}(1)=$

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Aicharya · Accepted Answer

$y=(1+x)\left(1+x^2ight)\left(1+x^4ight)\left(1+x^8ight)...(i)$ Taking 'log' on both sides, we get $\begin{aligned} & \log y=\log (1+x)+\log \left(1+x^2ight)+\log \left(1+x^4ight) \ &+ \log \left(1+x^8ight) \end{aligned}$ Differentiating w.r.t. $x$, we get $\frac{1 \mathrm{~d} y}{y}=\frac{1}{1+x}+\frac{2 x}{1+x^2}+\frac{4 x^3}{1+x^4}+\frac{8 x^7}{1+x^8}...(ii)$ At $x=1$, (i) $\Rightarrow y=16$ $	herefore \quad$ (ii) $\Rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} x}ight)_{x=1}=16\left(\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}ight)=120$

If $\mathrm{f}(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right)\left(1+x^8\right)$, then $f^{\prime}(1)=$

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