derivatives of natural logs and exponents

[noboost4you]

The problem I have is to find the derivative of the function:

f(x) = (ln x)^x

I know the derivative of ln x is 1/x, but the exponent is throwing me off. Can anyone offering any help? Thanks alot

[phoenixthoth]

a^{b}=e^{b\ln a}

so \left( \ln x\right) ^{x}=e^{x\ln \left( \ln x\right) }. now use product and chain rules.

[Ambitwistor]

Another (very similar) approach is to take the log of both sides before you take the derivative.. use the chain rule to write d(ln(f(x))/dx in terms of df/dx, and solve for df/dx.

[noboost4you]

Originally posted by phoenixthoth
a^{b}=e^{b\ln a}

so \left( \ln x\right) ^{x}=e^{x\ln \left( \ln x\right) }. now use product and chain rules.

Would I then have to substitute? I'm still not completely following... Thanks again

[phoenixthoth]

as long as your answer has only x's in it, it should be ok. you could simplify the e^{x\ln \left( \ln x\right) } back to \left( \ln x\right) ^{x} if you want.

[noboost4you]

would the derivative of (ln x)^x be:

xe^(1/x) ??

i don't know how to use the power and chain rule on e^xln(ln x)

[phoenixthoth]

we have y=e^{x\ln \left( \ln x\right) }.

this can be written as y=e^{u} where u=x\ln \left( \ln x\right) .

the chain rule is that \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}.

\frac{dy}{du}=e^{u}. to find \frac{du}{dx}, note that u is the
product of x and \ln \left( \ln x\right) .

\frac{du}{dx}=\left( \frac{d}{dx}x\right) \ln \left( \ln x\right) +x\frac{d
}{dx}\ln \left( \ln x\right) . \frac{d}{dx}x=1 and to find \frac{d}{dx}
\ln \left( \ln x\right) , it may be useful to write v=\ln w where w=\ln x
.

\frac{d}{dx}\ln \left( \ln x\right) =\frac{dv}{dx}=\frac{dv}{dw}\frac{dw}{dx
}.

\frac{dv}{dw}=\frac{1}{w} and \frac{dw}{dx}=\frac{1}{x}. hence \frac{d
}{dx}\ln \left( \ln x\right) =\frac{1}{w}\frac{1}{x}=\frac{1}{\ln x}\frac{1}{
x}=\frac{1}{x\ln x}.

putting this back into the most recent expression for \frac{du}{dx}, we
get \frac{du}{dx}=1\cdot \ln \left( \ln x\right) +x\left( \frac{1}{x\ln x}
\right) =\ln \left( \ln x\right) +\frac{1}{\ln x}.

putting this back into the most recent expression for \frac{dy}{dx}, we
get \frac{dy}{dx}=e^{u}\left( \ln \left( \ln x\right) +\frac{1}{\ln x}
\right) =e^{x\ln \left( \ln x\right) }\left( \ln \left( \ln x\right) +\frac{1
}{\ln x}\right) .

since \left( \ln x\right) ^{x}=e^{x\ln \left( \ln x\right) }, we get
\frac{dy}{dx}=e^{x\ln \left( \ln x\right) }\left( \ln \left( \ln x\right) +
\frac{1}{\ln x}\right) =\left( \ln x\right) ^{x}\left( \ln \left( \ln
x\right) +\frac{1}{\ln x}\right) . either the middle or right side of this equation may be acceptable.

[noboost4you]

I understand what you wrote, but I just can't figure out how you turned (ln x)^x into e^xln(ln x)

Please elaborate. Otherwise, everything else has been very helpful.

[phoenixthoth]

it's based on the property e^{\ln a}=a. if we raise both sides to the b power, we get \left( e^{\ln a}\right) ^{b}=a^{b} which becomes a^{b}=e^{b\ln a}. in this case, a=\ln x and b=x.