Natural Log Derivatives

Just use the chain rule.

i prefer an oscillating scale of logarithmic deviations myself

oscillating scale?
Never heard of this one. What is it?

Believe me, use the chain rule. Look it up in any calculus book.

Chain rule
Ok, so I looked up the chain rule. It says if r(x)=(mnp)(x)=m(n(p(x)) then r'(x)=m'(n(p(x))*n'(p(x)*p'(x).

Does this translate to: ln(lnx^2)^5 = ln(ln x^2)^5 * ln x^2 = 5ln(ln x^2)*2ln x = 5/2 ln x * 2/x = 10/x lnx ? or is it 10/x + lnx?

Or is it completely out of the ballpark?

Thanks!

Re: Chain rule

You stated it correctly, but then applied it incorrectly. It refers to chaining derivatives, not logarithms:

Since the derivative of ln(x) is 1/x,

ln((ln x^2)^5) = {1/[(ln x^2)^5]} * 5 ln(x^2)^4 * [1/(x^2)] * 2x

Now just simplify the terms.

I like to look at the chain rule as working from the inside out. IOW start with the function that is on the outside of all the parenthesis.

f(x)=ln((ln x^2)^5)=ln(g(x)) where g(x)=(ln x^2)^5
so f'(x)=1/g(x)*g'(x)

now work on the function g(x)
g(x)=h(x)^5 where h(x)=ln(x^2)
so
g'(x)=5h(x)^4*h'(x)

Now h(x)
h(x)=ln(i(x)) where i(x)=x^2
h'(x)=1/i(x)*i'(x)

Finally i(x)
i(x)=x^2
i'(x)=2x

Putting then all together
f(x)=ln((ln x^2)^5)=ln((ln(i(x))^5)=ln(h(x)^5)=ln(g(x)) and
f'(x)=[1/((ln(x^2)(^5][5(ln(x^2))^4][1/(x^2)][2x]

Now do a page of algebra and simplify, simplify. simplify.

Rap c∫;?/

Integrals
Thanks guys. I tried the problem again based on your info and got (the algebra way was too complicated for me):

5 ln(lnx^2) * 2 ln x = 5/ln x^2 *2/x = 10/ x+ lnx

or is it 10/x*lnx???

Thanks!

Maybe 10/ln(lnx)?

ln((ln x^2)^5) = {1/[(ln x^2)^5]} * 5 ln(x^2)^4 * [1/(x^2)] * 2x

Let's cancel the first two terms:

= [5/ln(x^2)]* [1/(x^2)] * 2x

= [5/ln(x^2)]* (2/x)

= 10/xln(x^2)

= 10/2xln(x)

= 5/xln(x)