How do you find the derivative of y= lnx^lnx?

Question

I need to learn this stuff. But I am so lost. If you can solve it, I'd appreciate it if you can provide the steps also.

Answer 1

I'm assuming you mean

y = (ln[x])^(ln[x])

This is a "function to a function" type question, which means you *have* to use logarithmic differentiation. This means taking the natural log (ln) of both sides, and then differentiating implicitly.

ln(y) = ln [(ln[x])^(ln[x])]

By the log property that ln(a^c) = c*ln(a), we apply this to the right hand side and get

ln(y) = ln(x) ln(ln(x))

Now, differentiate implicitly with respect to x. Don't forget to use the product rule and chain rule on the right hand side.

(1/y) (dy/dx) = (1/x) ln(ln(x)) + ln(x) [1/(ln(x))] [1/x]

Multiplying both sides by y,

dy/dx = y { (1/x) ln(ln(x)) + ln(x) [1/(ln(x))] [1/x] }

And now, replacing y = (lnx)^(lnx), we have

dy/dx = [ (lnx)^(lnx) ] { (1/x) ln(ln(x)) + ln(x) [1/(ln(x))] [1/x] }

This of course can be simplified, which I'll leave up to you to do. But these are the basic steps.

Answer 2

y = lnx^lnx
taking Natural log both sides we get.
lny = lnx ln(lnx)
Differentiating both sides with respect to x we get
(1/y) dy/dx = lnx (1/lnx)(1/x) + ln(lnx) /x
dy/dx = y{ lnx (1/lnx)(1/x) + ln(lnx) /x)}
dy/dx = lnx^lnx{ lnx (1/lnx)(1/x) + ln(lnx) /x)}

I hope it will help

Answer 3

y=ln(x)^ln(x)=e^(ln(x)*ln(x))=e^ln(x)^2
y'=2ln(x)*1/x*e^ln(x)^2=2ln(x)*ln(x)^ln(x)*1/x