Answers

2014-05-17T17:05:27+05:30
F(0) = a^(a+1)
f(x) = {(a+x) / (1+x)}^(a+1+2x)
take both side log
logf(x) = (a+1+2x)log{(a+x)/(1+x)}
take d/dx
f '(x)/f(x) = 2log{(a+x)/(1+x)} + {(a+1+2x)(1+x)(1-a)}/{(a+x)(1+x)^2}
f '(0) = f(0){2loga + (1+a)(1-a)/a}
f '(0) = a^(a+1){2loga + (1-a^2)/a}

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