Question

f(x) = {(a+x)/(1+x)}^(a+1+2x)
show that
f'(0) = a^ (a+1) {2loga + (1-a^2)/a}

Answer 1

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}