What is the first differential of $f(x) = 1/(x(1+lnx)$ ?

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Answer 1 · 2017-03-03T06:01:25

$-(2+lnx)/(x^2(1+lnx)^2$

$f(x) = 1/(x(1+lnx)$

$= (x*(1+lnx))^-1$

Applying the power rule and the chain rule:

$= -(x*(1+lnx))^-2 * d/dx(x*(1+lnx))$

$= -(x*(1+lnx))^-2 * x*(0+1/x)+(1+lnx)*1$
[Product rule and standard differential]

$= -(x*(1+lnx))^-2 * (1 +1 +lnx)$

$= -(2+lnx)/(x^2(1+lnx)^2$

What is the first differential of f(x) = 1/(x(1+lnx)f(x) = 1/(x(1+lnx) ?