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 a lot

 

[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.

 

[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 $$\ln \left( \ln x\right)$$.

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

$$\frac{d}{dx}\ln \left( \ln x\right) =\frac{dv}{dx}=\frac{dv}{dw}\frac{dw}{dx<br /> }.$$

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

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

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

since $$\left( \ln x\right) ^{x}=e^{x\ln \left( \ln x\right) }$$, we get $$\frac{dy}{dx}=e^{x\ln \left( \ln x\right) }\left( \ln \left( \ln x\right) +<br /> \frac{1}{\ln x}\right) =\left( \ln x\right) ^{x}\left( \ln \left( \ln<br /> x\right) +\frac{1}{\ln x}\right)$$. 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 $$b=x$$.