Mathematics
OpenStudy (anonymous):

use logarithmic differentiation to find y = x^(lnx)

myininaya (myininaya):

$\ln(y)=\ln(x^{\ln(x)})$

OpenStudy (anonymous):

you have a choice. myininaya will do it one way, i will do it another

myininaya (myininaya):

$\ln(y)=\ln(x) \ln(x)$

myininaya (myininaya):

$\ln(y)=[\ln(x)]^2$

myininaya (myininaya):

$\frac{y'}{y}=2 \ln(x) \frac{1}{x}$

myininaya (myininaya):

now multiply y on both sides

OpenStudy (anonymous):

myininya did u do product rule?

myininaya (myininaya):

chain rule

OpenStudy (anonymous):

$x^{\ln(x)}=e^{\ln(x)\times \ln(x)}$so $\frac{d}{dx}x^{\ln(x)}=\frac{d}{dx}e^{\ln(x)\times \ln(x)}$ $=e^{\ln(x)\times \ln(x)}\times \frac{d}{dx}\ln(x)\times \ln(x)$ and $\frac{d}{dx}\ln^2(x)=2\ln(x)\frac{1}{x}$ so your final answer is $=e^{\ln(x)\times \ln(x)}\times \frac{2\ln(x)}{x}$

OpenStudy (anonymous):

why did u multiply by 1/x?

OpenStudy (anonymous):

chain rule. the derivative of $f^2(x)$ is $2f(x)f'(x)$

OpenStudy (anonymous):

$\frac{d}{dx}\ln^2(x)=2\ln(x)\ln'(x)=2\ln(x)\frac{1}{x}$

OpenStudy (anonymous):

thamk you!

OpenStudy (anonymous):

when you see $f(x)^{g(x)}$ and you want the derivative, you have a choice. you can 1) take the log 2) take the derivative of the log 3) multiply the result by the original function

OpenStudy (anonymous):

or, you can rewrite $f(x)^{g(x)}=e^{g(x)\ln(f(x))}$ and take the derivative using the chain rule

OpenStudy (anonymous):

it amounts to the same thing because all the work is finding the derivative of $g(x)\ln(f(x))$