If y=(1+x)(1+x^{2})(1+x^{4}),then frac{dy}{dx | vedclass

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(A) Given $y=(1+x)(1+x^{2})(1+x^{4})$.We can write $y = \frac{(1-x)(1+x)(1+x^{2})(1+x^{4})}{1-x} = \frac{1-x^{8}}{1-x}$ for $x 
eq 1$.Using the quoti

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$28$

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(A) Given $y=(1+x)(1+x^{2})(1+x^{4})$.We can write $y = \frac{(1-x)(1+x)(1+x^{2})(1+x^{4})}{1-x} = \frac{1-x^{8}}{1-x}$ for $x 
eq 1$.Using the quotient rule,$\frac{dy}{dx} = \frac{(1-x)(-8x^{7}) - (1-x^{8})(-1)}{(1-x)^{2}} = \frac{-8x^{7}+8x^{8}+1-x^{8}}{(1-x)^{2}} = \frac{7x^{8}-8x^{7}+1}{(1-x)^{2}}$.Applying $L$'Hopital's rule or expanding the series,we evaluate the limit as $x 	o 1$.Alternatively,using logarithmic differentiation: $\ln y = \ln(1+x) + \ln(1+x^{2}) + \ln(1+x^{4})$.$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \frac{4x^{3}}{1+x^{4}}$.At $x=1$,$y = (2)(2)(2) = 8$.$\frac{1}{8} \frac{dy}{dx} = \frac{1}{2} + \frac{2}{2} + \frac{4}{2} = \frac{1}{2} + 1 + 2 = 3.5$.$\frac{dy}{dx} = 8 	imes 3.5 = 28$.

If $y=(1+x)(1+x^{2})(1+x^{4})$,then $\frac{dy}{dx}$ at $x=1$ is

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