If f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8),then f^{pr | vedclass

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(D) Given $f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$.Taking the natural logarithm on both sides:$\ln(f(x)) = \ln(1+x) + \ln(1+x^2) + \ln(1+x^4) + \ln(1+x^8)$.D

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(D) Given $f(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)$.Taking the natural logarithm on both sides:$\ln(f(x)) = \ln(1+x) + \ln(1+x^2) + \ln(1+x^4) + \ln(1+x^8)$.Differentiating both sides with respect to $x$:$\frac{f^{\prime}(x)}{f(x)} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8}$.At $x=1$,$f(1) = (1+1)(1+1)(1+1)(1+1) = 2 	imes 2 	imes 2 	imes 2 = 16$.Substituting $x=1$ into the derivative expression:$\frac{f^{\prime}(1)}{f(1)} = \frac{1}{1+1} + \frac{2(1)}{1+1^2} + \frac{4(1)^3}{1+1^4} + \frac{8(1)^7}{1+1^8} = \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} = \frac{1+2+4+8}{2} = \frac{15}{2}$.Therefore,$f^{\prime}(1) = f(1) 	imes \frac{15}{2} = 16 	imes \frac{15}{2} = 8 	imes 15 = 120$.

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

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