tricky derivative of natural log function

[chelle2007]

Hello all!

I always second guess myself with math. Can you tell me if I am headed in the correct direction or not>? Thank you for any/all replies!

Determine the derivative of f(x) = (ln x)^(ln x)

I am honestly stuck. In my textbook we have practice problems where we take the natural log of the function but natural logs already exist in the function.

I tried:

since x = e^(lnx)

then subbing x gives me: ln(e^lnx)^lnx

and that simplifies it to f(x) = lnx * lnx

Help?!?!

 

[tkhunny]

Try this: ln(f) = ln(x)*ln(ln(x)).

It's pretty deep, but let's see what you get from there. Find df/dx.

 

[pka]

Determine the derivative of f(x) = (ln x)^(ln x)

\(\displaystyle \ln(f(x))=\ln(n)\cdot\ln((\ln(x)))\)

 

[Deleted member 4993]

Hello all!

I always second guess myself with math. Can you tell me if I am headed in the correct direction or not>? Thank you for any/all replies!

Determine the derivative of f(x) = (ln x)^(ln x)

I am honestly stuck. In my textbook we have practice problems where we take the natural log of the function but natural logs already exist in the function.

I tried:

since x = e^(lnx)

then subbing x gives me: [ln(e^lnx)]^lnx


Help?!?!

However - I would do it following way:

u = ln(x)

du/dx = 1/x

f(u) = u^u

ln[f] = u * ln(u)

1/f * df/du = ln(u) + 1

df/du = u^u * (ln(u) + 1)

then

df/dx = df/du * du/dx

Now do the algebra and simplify (if possible)......

 

[chelle2007]

Try this: ln(f) = ln(x)*ln(ln(x)).

It's pretty deep, but let's see what you get from there. Find df/dx.

That was the other way my textbook described. logarithmic differentiation right?

This is what I did:

ln f(x) = ln (lnx) ^ (ln x) = lnx * ln(ln(x))

product rule to get:

1/x * ln(ln(x))+ ln(x) * (1/ln(x)) = (ln(ln(x)))/x +1

this times the original functions gives me the derivative:

[(ln(ln(x)))/x +1] * (ln(x)^ln(x))

Right? :-(

 

[soroban]

Hello, chelle2007!

That was the other way my textbook described. .Logarithmic differentiation, right?

This is what I did: ..\(\displaystyle \ln f \:=\: \ln\big[(\ln x)\big]^{\ln x} \:=\: \ln x\cdot \ln(\ln x)\)

Product rule to get: ..\(\displaystyle \dfrac{f'}{f} \:=\:\dfrac{1}{x}\cdot\ln(\ln x)+ \ln x \cdot\dfrac{1}{\ln x}\) . . . . no

Product Rule: .\(\displaystyle \dfrac{f'}{f} \;=\;\dfrac{1}{x}\!\cdot\!\ln(\ln x) + \ln x\!\cdot\!\dfrac{1}{\ln x}\!\cdot\!\dfrac{1}{x}\)

. . . . . . . . . . .\(\displaystyle \dfrac{f'}{f} \;=\;\dfrac{1}{x}\!\cdot\!\ln(\ln x) + \dfrac{1}{x}\)

. . . . . . . . . . .\(\displaystyle \dfrac{f'}{f} \;=\;\dfrac{\ln(\ln x) + 1}{x}\)

. . . . . . . . . . .\(\displaystyle f' \;=\;\dfrac{(\ln x)^{\ln x}[\ln(\ln x) + 1]}{x}\)

 

[chelle2007]

Thank you for everyone's replies! I worked out both ways and got the same answer as the last reply! Thank you soroban for goin step by step. It's been a while since I last did any kind of calculus!