What is the derivative of $f (x) = \frac{1}{x ln x}$ $f(x) = 1/(xlnx)$ ?

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Answer 1 · 2018-03-22T14:18:23

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

We know that,
$color(red)((1) d/(dx)(X^n)=nX^(n-1)$
$color(red)((2) d/(dx)(u*v)=u*(dv)/(dx)+v*(du)/(dx)$

Here,
$f(x)=1/(xlnx)=(xlnx)^-1$

Using chain rule and (1)

$f^'(x)=-1*(xlnx)^(-1-1)d/(dx)(xlnx)$

$=-(xlnx)^-2[xd/(dx)(lnx)+lnxd/(dx)(x)],..to$ Apply(2)

$=-1/(xlnx)^2[x*1/x+lnx(1)]$

$=-1/(xlnx)^2[1+lnx]$

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

What is the derivative of f(x) = 1/(xlnx)f(x)=1xlnxf(x) = 1/(xlnx)?