Question

Find the slope at any value of x,if y=(lnx)^1/lnx ?

Answer 1

`y'=(ln(x))^(1/ln(x))*(-ln(ln(x))/(ln(x))^2x)+1/(x*(ln(x))^2))`

Explanation:

Taking the logarithm on both sides:
`ln(y)=(ln(x))^(-1)ln(ln(x))`
Differentiating with respect to `x'`:
`(y')/y=(-1)(ln(x))^(-2)1/xln(ln(x))+ln(x)^(-1)+1/ln(x)*1/x`
so we get
`y'=(ln(x))^(1/ln(x))*(-ln(ln(x))/(ln(x))^2*x)+1/(x*ln(x))^2)`