# Does anyone know how to find the derivative of y= ln(lnx^2)?I tried to do everything, but I don't know how to start it. Please help me.

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ln(lnx^2)

This is just chain rule.

`(1)/(lnx^2)xx(1)/(x^2)xx(2x)`

``After that its a matter of simplifying if you want to. You'll have to remember log rules. The two and the x will cancel to leave you with 1/xlnx.

We'll find the derivative of a composed function.

y=ln(lnx^2), where lnx^2=t(u(x)) and u(x)=x^2

y=ln(t(u(x)))

y'=[ln(t(u(x)))]'=[1/t(u(x))]*[t(u(x))]'

t(u(x))'=(1/x^2)*(x^2)'=(1/x^2)*(2*x)=2/x

y'=[ln(t(u(x)))]'=(1/lnx^2)*(2/x)=1/(1/2)*lnx^2x=1/lnx^x

**y'=1/lnx^x**

y= ln(lnx^2)

To find the derivative of the y.

We use thr results : d/dx(x^n) = nx^(n-1) and

d/dx { f (g(x))} ={ f(g(x)}' = { f '(g(x))}*g'(x), the chain rule formula.

Therefore d/dx{ln(f(x)} ={ lnf(x)}' = {1/f(x)}f'(x).

Coming to our question,

y = ln(lnx^2). To find the dy/dx:

dy/dx = (ln(lnx^2))'. Using the above chain rule formula we get:

dy/dx = {1/(lnx^2)} (lnx^2)'

= {1/(lnx^2)} (1/x^2)(x^2)'

=(1/(lnx^2)(1/x^2)(2x)

=2/{xlnx^2)

=1/(xlnx)