Find the derivative of the function given by | vedclass

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(A) Given the function $f(x)=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})$.Taking the natural logarithm on both sides,we get:$\log f(x) = \log(1+x) + \log(1+x^{2}

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(A) Given the function $f(x)=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})$.Taking the natural logarithm on both sides,we get:$\log f(x) = \log(1+x) + \log(1+x^{2}) + \log(1+x^{4}) + \log(1+x^{8})$.Differentiating both sides with respect to $x$,we obtain:$\frac{1}{f(x)} \cdot f^{\prime}(x) = \frac{d}{dx} \log(1+x) + \frac{d}{dx} \log(1+x^{2}) + \frac{d}{dx} \log(1+x^{4}) + \frac{d}{dx} \log(1+x^{8})$.Using the chain rule,we have:$\frac{f^{\prime}(x)}{f(x)} = \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \frac{4x^{3}}{1+x^{4}} + \frac{8x^{7}}{1+x^{8}}$.Thus,$f^{\prime}(x) = f(x) \left[ \frac{1}{1+x} + \frac{2x}{1+x^{2}} + \frac{4x^{3}}{1+x^{4}} + \frac{8x^{7}}{1+x^{8}} ight]$.Now,to find $f^{\prime}(1)$,we substitute $x=1$:$f(1) = (1+1)(1+1^{2})(1+1^{4})(1+1^{8}) = 2 	imes 2 	imes 2 	imes 2 = 16$.$f^{\prime}(1) = f(1) \left[ \frac{1}{1+1} + \frac{2(1)}{1+1^{2}} + \frac{4(1)^{3}}{1+1^{4}} + \frac{8(1)^{7}}{1+1^{8}} ight]$.$f^{\prime}(1) = 16 \left[ \frac{1}{2} + \frac{2}{2} + \frac{4}{2} + \frac{8}{2} ight]$.$f^{\prime}(1) = 16 \left[ \frac{1+2+4+8}{2} ight] = 16 	imes \frac{15}{2} = 8 	imes 15 = 120$.

Find the derivative of the function given by $f(x)=(1+x)(1+x^{2})(1+x^{4})(1+x^{8})$ and hence find $f^{\prime}(1).$

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